CODEFORCES-PROBLEMSET

1A

水题   然而看不仔细爆int了

c++

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    double n, m, a;
    cin >> n >> m >> a;
    ll ans = ll(ceil(n / a)) * ll(ceil(m / a));
    cout << ans << endl;
    return 0;
}

py3

import math
t = input().split()
n, m, a = map(int, t)
print(int(math.ceil(n / a) * math.ceil(m / a)))

1B

有毒的模拟水题......

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    while(n--){
        string s;
        cin >> s;
        bool flag = false;
        int i = 0;
        while(i < s.size() && 'A' <= s[i] && s[i] <= 'Z') ++i;
        for(; i < s.size(); ++i){
            if('A' <= s[i] && s[i] <= 'Z'){
                flag = true;
                break;
            }
        }
        if(flag){
            int c = 0;
            int index = s.find('C');
            string ans = "";
            for(int i = index + 1; i < s.size(); ++i) c = 10 * c + s[i] - '0';
            while(c){
                int k = c % 26;
                if(k == 0){
                    ans += 'Z';
                    --c;
                }
                else ans += ('A' + k - 1);
                c /= 26;
            }
            reverse(ans.begin(), ans.end());
            cout << ans << s.substr(1, index-1) << endl;
        }
        else{
            int index = 0;
            string ans = "R";
            int r = 0;
            for(; index < s.size(); ++index){
                if('0' <= s[index] && s[index] <= '9') break;
                r = 26 * r + (s[index] - 'A' + 1);
            }
            ans += s.substr(index, s.size());
            ans += 'C';
            cout << ans << r << endl;
        }
    }
    return 0;
}

1C

mdzz。。。

余弦定理求三角形内角，也是圆上的圆周角，乘以2得圆心角

求三圆心角最大公约数得正多边形每一份的圆心角ang，2*pi/ang得出边数n

海伦公式+正弦定理得三角形外接圆半径r = (a * b * c) / (4 * s)

s = n * r * r * sin(ang) / 2

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-2;
double fgcd(double a, double b){
     if (fabs(a) < eps)
         return b;
     if (fabs(b) < eps)
         return a;
     return fgcd(b, fmod(a, b));
}
double diameter(double a, double b, double c) {
    double p = (a + b + c) / 2;
    double s = sqrt(p * (p - a) * (p - b) * (p - c));
    return a * b * c / (4 *s);
}
int main() {
    double x[3], y[3], line[3];
    for(int i = 0; i < 3; ++i) scanf("%lf %lf", &x[i], &y[i]);
    for(int i = 0; i < 3; ++i)
        line[i] = sqrt((x[i] - x[(i+1)%3]) * (x[i] - x[(i+1)%3]) + (y[i] - y[(i+1)%3]) * (y[i] - y[(i+1)%3]));
    double r = diameter(line[0], line[1], line[2]);
    double angle[3];
    for(int i = 0; i < 2; ++i)
        angle[i] = acos(1 - line[i] * line[i] / (2 * r * r));
    angle[2] = 2 * acos(-1) - angle[0] - angle[1];
    double ang = fgcd(angle[0], fgcd(angle[1], angle[2]));
    printf("%.6f
", r * r * sin(ang) / 2 * (2 * acos(-1) / ang));
    return 0;
}

2A

模拟水题   略麻烦(说到底是思路不清淅不严谨+弱)

#include <bits/stdc++.h>
using namespace std;

struct info{
    string name;
    int score;
}aaa[1111];
map<string, int> game;
int main(){
    int n;
    cin >> n;
    string name, ans;
    int score;
    for(int i = 0; i < n; ++i){
        cin >> name >> score;
        aaa[i].name = name;
        aaa[i].score = score;
        game[name] += score;
    }
    int maxScore = -1;
    for(int i = 0; i < n; ++i) maxScore = max(maxScore, game[aaa[i].name]);
    map<string, int> tmp;
    for(int i = 0; i < n; ++i){
        if(game[aaa[i].name] < maxScore) continue;
        tmp[aaa[i].name] += aaa[i].score;
        if(tmp[aaa[i].name] >= maxScore){
            ans = aaa[i].name;
            break;
        }
    }
    cout << ans << endl;
    return 0;
}

2B

然后就2B了.....

求2最少几个5最少几个然后min(num_two, num_five)就是答案   剩下输出路径

#include <bits/stdc++.h>
using namespace std;

int matrix[1111][1111][2];
int dp[1111][1111][2];
int step[1111][1111][2];

void print(int nx, int ny, int k){
    if(nx == 1 && ny == 1) return ;
    if(step[nx][ny][k]){
        print(nx-1, ny, k);
        printf("D");
    }
    else{
        print(nx, ny-1, k);
        printf("R");
    }
}
int main(){
    int n;
    scanf("%d", &n);
    int zerox = 0, zeroy = 0;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j){
            int k;
            scanf("%d", &k);
            if(k == 0){
                zerox = i;
                zeroy = j;
                continue;
            }
            int numtwo = 0, numfive = 0;
            while(!(k & 1)){
                ++numtwo;
                k >>= 1;
            }
            while(k % 5 == 0){
                ++numfive;
                k /= 5;
            }
            matrix[i][j][0] = numtwo;
            matrix[i][j][1] = numfive;
        }
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[1][1][0] = matrix[1][1][0];
    dp[1][1][1] = matrix[1][1][1];
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j){
            if(i == 1 && j == 1) continue;
            for(int k = 0; k < 2; ++k){
                dp[i][j][k] = matrix[i][j][k] + min(dp[i-1][j][k], dp[i][j-1][k]);
                if(dp[i-1][j][k] < dp[i][j-1][k]) step[i][j][k] = 1;
            }
        }
    }
    int ans = min(dp[n][n][0], dp[n][n][1]);
    if(ans == 0) puts("0");
    else if(zerox && zeroy){
        puts("1");
        int nx = 1, ny = 1;
        while(nx < zerox) ++nx, printf("D");
        while(ny < zeroy) ++ny, printf("R");
        while(nx < n) ++nx, printf("D");
        while(ny < n) ++ny, printf("R");
        return 0;
    }
    else printf("%d
", ans);
    int k = dp[n][n][0] < dp[n][n][1] ? 0 : 1;
    print(n, n, k);
    return 0;
}

2C

模拟退火   先放着

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
#define _pow(a) ((a)*(a))
const double eps = 1e-6;

struct circle{ double x, y, r; }c[3];
double angle[3];
int cx[] = {0, 0, 1, -1}, cy[] = {1, -1, 0, 0};

double dis(double x1, double y1, double x2, double y2){
    return sqrt(_pow(x1 - x2) + _pow(y1 - y2));
}

double calc(double x, double y){
    for(int i = 0; i < 3; ++i) angle[i] = dis(x, y, c[i].x, c[i].y) / c[i].r;
    double value = 0;
    for(int i = 0; i < 3; ++i) value += _pow(angle[i] - angle[(i + 1) % 3]);
    return value;
}

int main(){
    double x = 0, y = 0;
    for(int i = 0; i < 3; ++i){
        scanf("%lf %lf %lf", &c[i].x, &c[i].y, &c[i].r);
        x += c[i].x / 3.0;
        y += c[i].y / 3.0;
    }
    double deviation = calc(x, y);
    double step = 1;
    for(int i = 0; i <= 1e5; ++i){
        bool flag = false;
        double xx, yy;
        for(int j = 0; j < 4; ++j){
            double nx = x + cx[j] * step, ny = y + cy[j] * step;
            double nDeviation = calc(nx, ny);
            if(nDeviation < deviation){
                deviation = nDeviation;
                flag = true;
                xx = nx;
                yy = ny;
            }
        }
        if(flag){
            x = xx;
            y = yy;
        }
        else step /= 2.0;
    }
    if(deviation <= eps) printf("%.5f %.5f
", x, y);
    return 0;
}

3A

水题  但是大神的代码总是让我amazing，简单清晰直接

#include <bits/stdc++.h>
using namespace std;

int main(){
    char s[3], t[3];
    scanf("%s %s", s, t);
    int a = s[0] - t[0], b = s[1] - t[1];
    char c = a > 0 ? 'L' : (a = -a, 'R');
    char d = b > 0 ? 'D' : (b = -b, 'U');
    printf("%d", a > b ? a : b);
    while(a || b){
        puts("");
        if(a) --a, putchar(c);
        if(b) --b, putchar(d);
    }
    return 0;
}

3B

分开按capacity排序  再维护前缀和  再枚举space为1的船的数量

代码好丑...........

#include <bits/stdc++.h>
using namespace std;

struct vihecle{
    int id;
    int cpa;
}p1[111111], p2[111111];
int presum1[111111], presum2[111111];

bool cmp(vihecle a, vihecle b){
    return a.cpa > b.cpa;
}

int main(){
    int n, v;
    cin >> n >> v;
    int i1 = 0, i2 = 0;
    presum1[0] = presum2[0] = 0;
    for(int i = 0; i < n; ++i){
        int t, pp;
        cin >> t >> pp;
        if(t == 1){
            p1[i1].cpa=pp;
            p1[i1++].id=i+1;
        }
        else{
            p2[i2].cpa=pp;
            p2[i2++].id=i+1;
        }
    }
    sort(p1, p1 + i1, cmp);
    sort(p2, p2 + i2, cmp);
    for(int i = 1; i <= i1; ++i) presum1[i] = presum1[i-1] + p1[i-1].cpa;
    for(int i = 1; i <= i2; ++i) presum2[i] = presum2[i-1] + p2[i-1].cpa;
    int ans = 0, ii1 = 0, ii2 = 0;
    for(int i = 0; i <= i1; ++i){
        if(i > v) break;
        int tmp = presum1[i] + presum2[min((v-i)/2, i2)];
        if(tmp > ans){
            ans = tmp;
            ii1 = i;
            ii2 = min((v - i) / 2, i2);
        }
    }
    cout << ans << endl;
    for(int i = 0; i < ii1; ++i) cout << p1[i].id << " ";
    for(int i = 0; i < ii2; ++i) cout << p2[i].id << " ";
    return 0;
}

3C

恶心的模拟..........讲真......玩起来容易写起来恶心.......

#include <bits/stdc++.h>
using namespace std;

char g[5][5];
int c1, c2;
bool f1, f2;

int main(){
    for(int i = 0; i < 3; ++i) scanf("%s", g[i]);
    c1 = c2 = 0;
    f1 = f2 = false;

    for(int i = 0; i < 3; ++i)
        if(g[i][0] == g[i][1] && g[i][1] == g[i][2])
            if(g[i][0] == 'X') f1 = true;
            else if(g[i][0] == '0') f2 = true;
    for(int i = 0; i < 3; ++i)
        if(g[0][i] == g[1][i] && g[1][i] == g[2][i])
            if(g[0][i] == 'X') f1 = true;
            else if(g[0][i] == '0') f2 = true;
    if(g[0][0] == g[1][1] && g[1][1] == g[2][2])
        if(g[0][0] == 'X') f1 = true;
        else if(g[0][0] == '0') f2 = true;
    if(g[0][2] == g[1][1] && g[1][1] == g[2][0])
        if(g[1][1] == 'X') f1 = true;
        else if(g[1][1] == '0') f2 = true;

    for(int i = 0; i < 3; ++i)
        for(int j = 0; j < 3; ++j)
            if(g[i][j] == 'X') ++c1;
            else if(g[i][j] == '0') ++c2;
    if(c1 == c2 || c1 == c2 + 1){
        if(f1 && f2) puts("illegal");
        else if(f1 && !f2){
            if(c1 == c2 + 1) puts("the first player won");
            else puts("illegal");
        }
        else if(!f1 && f2){
            if(c1 == c2) puts("the second player won");
            else puts("illegal");
        }
        else{
            if(c1 + c2 == 9) puts("draw");
            else if(c1 + c2 < 9){
                if(c1 == c2) puts("first");
                else if(c1 == c2 + 1) puts("second");
            }
            else puts("illegal");
        }
    }
    else puts("illegal");
    return 0;
}

3D

对于处理中每一个过程都有'('不少于')'，将'?'置为')'，若'('少于')'则在前面的'?'中拿出转换价值最小的进行转换

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    string s;
    cin >> s;
    ll cnt = 0, ans = 0;
    priority_queue<pair<int, int> > q;
    for(int i = 0; i < s.size(); ++i){
        if(s[i] == '(') ++cnt;
        else if(s[i] == ')') --cnt;
        else{
            int a, b;
            cin >> a >> b;
            ans += b;
            --cnt;
            s[i] = ')';
            q.push(make_pair(b - a, i));
        }
        if(cnt < 0){
            if(q.empty()) break;
            pair<int, int> p = q.top();
            q.pop();
            ans -= p.first;
            s[p.second] = '(';
            cnt += 2;
        }
    }
    if(cnt) cout << "-1" << endl;
    else cout << ans << endl << s << endl;
    return 0;
}

4A

超级水......

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    cout << (((n & 1) || n == 2) ? "NO" : "YES") << endl;
    return 0;
}

4B

水，随便贪一下

#include <bits/stdc++.h>
using namespace std;

struct {
    int mini, maxi;
}p[33];

int main(){
    int d, maxSum;
    cin >> d >> maxSum;
    int totalMin = 0, totalMax = 0;
    for(int i = 0; i < d; ++i) cin >> p[i].mini >> p[i].maxi, totalMin += p[i].mini, totalMax += p[i].maxi;
    if(totalMin > maxSum || totalMax < maxSum)
        cout << "NO" << endl;
    else{
        cout << "YES" << endl;
        int t = maxSum - totalMin;
        for(int i = 0; i < d; ++i){
            int ansi = p[i].mini;
            if(t){
                ansi = min(p[i].maxi, ansi + t);
                t -= (ansi - p[i].mini);
            }
            cout << ansi << " 
"[i == d-1];
        }
    }
    return 0;
}

4C

水，输入保证只有小写字母，map水过

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    map<string, int> name;
    while(n--){
        string s;
        cin >> s;
        if(name[s] == 0) cout << "OK" << endl;
        else cout << s << name[s] << endl;
        ++name[s];
    }
    return 0;
}

4D

n^2的暴力dp.......

#include <bits/stdc++.h>
using namespace std;

int n;
struct env{
    int w, h;
}p[5555];
int dp[5555], pre[5555];

int dfs(int k){
    if(dp[k]) return dp[k];
    for(int i = 1; i <= n; ++i){
        if(p[k].w < p[i].w && p[k].h < p[i].h){
            if(dfs(i) + 1 > dp[k]){
                pre[k] = i;
                dp[k] = dfs(i) + 1;
            }
        }
    }
    return dp[k];
}

int main(){
    cin >> n;
    for(int i = 0; i <= n; ++i)
        cin >> p[i].w >> p[i].h;
    cout << dfs(0) << endl;
    for(int i = pre[0]; i > 0; i = pre[i]) cout << i << " ";
    cout << endl;
    return 0;
}

5A

水

#include <bits/stdc++.h>
using namespace std;

int main(){
    string s;
    int ans = 0, member = 0;
    while(getline(cin, s)){
        if(s[0] == '+') ++member;
        else if(s[0] == '-') --member;
        else{
            int len = s.end() - find(s.begin(), s.end(), ':') - 1;
            ans += member * len;
        }
    }
    cout << ans << endl;
    return 0;
}

5B

一眼样例明白大概但是奇数的情况要交替........

#include <bits/stdc++.h>
using namespace std;

int main(){
    vector<string> v;
    string s;
    int len = 0;
    while(getline(cin, s)){
        v.push_back(s);
        len = max(len, int(s.size()));
    }
    cout << string(len+2, '*') << endl;
    int k = -1;
    for(vector<string>::iterator i = v.begin(); i != v.end(); ++i){
        int l = len - (*i).size();
        string fr = string(floor(double(l)/2), ' '), ba = string(ceil(double(l)/2), ' ');
        if(l & 1){
            if(!k) cout << "*" << ba << *i << fr << "*" << endl;
            else cout << "*" << fr << *i << ba << "*" << endl;
            k = ~k;
        }
        else cout << "*" << fr << *i << ba << "*" << endl;
    }
    cout << string(len+2, '*') << endl;
    return 0;
}

5C

dp，让我智障好久的题........

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e6 + 7;
int dp[maxn];

int main(){
    ios::sync_with_stdio(false);
    memset(dp, 0, sizeof(dp));
    string s;
    cin >> s;
    stack<int> pos;
    while(!pos.empty()) pos.pop();
    int len = 0, sum = 1;
    for(int i = 0; i < s.size(); ++i){
        if(s[i] == '(') pos.push(i);
        else if(!pos.empty()){
            int j = pos.top();
            pos.pop();
            int tmpSize = i - j + 1;
            dp[i] = tmpSize + (j ? dp[j-1] : 0);
            if(dp[i] > len){
                len = dp[i];
                sum = 1;
            }
            else if(dp[i] == len) ++sum;
        }
    }
    cout << len << " " << sum << endl;
    return 0;
}

5D

运动学卧槽......放弃

分类还是渣的不成样子.....

#include <bits/stdc++.h>
using namespace std;

const double eps = 1e-8;
double a, v, l, d, w, t1, t2, vd, vt, r;

int fcmp(double x){
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

int main(){
    scanf("%lf %lf %lf %lf %lf", &a, &v, &l, &d, &w);
    vt = sqrt(w * w / 2 + a * d);
    r = l - d;
    if(fcmp(d - v * v / 2 / a) <= 0 && fcmp(sqrt(2 * a * d) - w) <= 0){
        t1 = sqrt(2 * d / a);
        vd = sqrt(2 * a * d);
    }
    else if(fcmp(d - v * v / 2 / a) >= 0 && fcmp(w - v) >= 0){
        t1 = d / v + v / 2 / a;
        vd = v;
    }
    else if(fcmp(v - vt) >= 0 && fcmp(vt - w) >= 0){
        t1 = (2 * vt - w) / a;
        vd = w;
    }
    else{
        t1 = d / v + (2 * v - w) / a - (2 * v * v - w * w) / 2 / a / v;
        vd = w;
    }
    if(fcmp(r - (v * v - vd * vd) / 2 / a) >= 0)
        t2 = r / v + (v - vd) / a - (v * v - vd * vd) / 2 / a / v;
    else
        t2 = (sqrt(2 * a * r + vd * vd) - vd) / a;
    printf("%.10f
", t1 + t2);
    return 0;
}

5E

找每个点左右分别严格高于这个点的点，有点像并查集的样子

还需要找处在中间的相同高度的点

每个点与左右两个高点及相同高度的点可以互相看到

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1e6 + 7;
int height[maxn], l[maxn], r[maxn], same[maxn];
ll ans = 0;

int main(){
    int n;
    scanf("%d", &n);
    int maxi = -1, pos = 0;
    for(int i = 0; i < n; ++i){
        scanf("%d", &height[i]);
        if(height[i] > maxi){
            maxi = height[i];
            pos = i;
        }
    }
    rotate(height, height + pos, height + n);
    height[n] = maxi;
    same[n] = 0;
    for(int i = n - 1; i >= 0; --i){
        r[i] = i + 1;
        while(r[i] < n && height[i] > height[r[i]]) r[i] = r[r[i]];
        if(r[i] != n && height[i] == height[r[i]]){
            same[i] = same[r[i]] + 1;
            r[i] = r[r[i]];
        }
    }
    for(int i = 1; i <= n; ++i){
        l[i] = i - 1;
        while(l[i] && height[i] >= height[l[i]]) l[i] = l[l[i]];
    }
    for(int i = 1; i < n; ++i){
        ans += (2 + same[i]);
        if(l[i] == 0 && r[i] == n) --ans;
    }
    printf("%I64d
", ans);
    return 0;
}

6A

水题

#include <bits/stdc++.h>
using namespace std;

int a, b, c, d;

void init(){
    int arr[4];
    cin >> arr[0] >> arr[1] >> arr[2] >> arr[3];
    sort(arr, arr + 4);
    a = arr[0]; b= arr[1]; c = arr[2]; d = arr[3];
}
int main(){
    init();
    if(a + b > c || a + b > d || a + c > d || b + c > d) puts("TRIANGLE");
    else if(a + b == c || a + b == d || a + c == d || b + c == d) puts("SEGMENT");
    else puts("IMPOSSIBLE");
    return 0;
}

6B

两层广搜过的，然而智障了......用一层广搜+set就行了

两层的辣鸡代码....:

#include <bits/stdc++.h>
using namespace std;

const int maxsize = 111;
int cx[] = {1, -1, 0, 0}, cy[] = {0, 0, 1, -1};
struct pos{
    int x, y;
    pos(int xx = 0, int yy = 0): x(xx), y(yy){};
};

char room[maxsize][maxsize];
bool flag[maxsize][maxsize];

int main(){
    int n, m;
    char color;
    cin >> n >> m;
    cin.get();
    cin >> color;
    queue<pos> mainpos;
    memset(flag, false, sizeof(flag));
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j){
            cin >> room[i][j];
            if(room[i][j] == color){
                mainpos.push(pos(i, j));
                flag[i][j] = true;
            }
        }
    int ans = 0;
    while(!mainpos.empty()){
        int x = mainpos.front().x, y = mainpos.front().y;
        mainpos.pop();
        for(int i = 0; i < 4; ++i){
            int nx = x + cx[i], ny = y + cy[i];
            if(nx < 0 || n <= nx || ny < 0 || m <= ny || flag[nx][ny] || room[nx][ny] == '.') continue;
            ++ans;
            queue<pos> tpos;
            tpos.push(pos(nx, ny));
            flag[nx][ny] = true;
            char tcolor = room[nx][ny];
            while(!tpos.empty()){
                int xx = tpos.front().x, yy = tpos.front().y;
                tpos.pop();
                for(int j = 0; j < 4; ++j){
                    int nxx = xx + cx[j], nyy = yy + cy[j];
                    if(nxx < 0 || n <= nxx || nyy < 0 || m <= nyy || room[nxx][nyy] != tcolor || flag[nxx][nyy]) continue;
                    flag[nxx][nyy] = true;
                    tpos.push(pos(nxx, nyy));
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}

一层 + set：

#include <bits/stdc++.h>
using namespace std;

const int maxsize = 111;
int cx[] = {1, -1, 0, 0}, cy[] = {0, 0, 1, -1};
struct pos{
    int x, y;
    pos(int xx = 0, int yy = 0): x(xx), y(yy){};
};

char room[maxsize][maxsize];

int main(){
    int n, m;
    char color;
    cin >> n >> m;
    cin.get();
    cin >> color;
    queue<pos> mainpos;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j){
            cin >> room[i][j];
            if(room[i][j] == color)
                mainpos.push(pos(i, j));
        }
    set<char> ans;
    while(!mainpos.empty()){
        int x = mainpos.front().x, y = mainpos.front().y;
        mainpos.pop();
        for(int i = 0; i < 4; ++i){
            int nx = x + cx[i], ny = y + cy[i];
            if(nx < 0 || n <= nx || ny < 0 || m <= ny || room[nx][ny] == color || room[nx][ny] == '.') continue;
            ans.insert(room[nx][ny]);
        }
    }
    cout << ans.size() << endl;
    return 0;
}

6C

水题，第一次用deque

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    deque<int> chb;
    for(int i = 0; i < n; ++i){
        int tmp;
        cin >> tmp;
        chb.push_back(tmp);
    }
    int t1 = chb.front(), ans1 = 1, t2 = 0, ans2 = 0;
    chb.pop_front();
    while(!chb.empty()){
        if(t1 <= t2){
            t1 += chb.front();
            ++ans1;
            chb.pop_front();
        }
        else{
            t2 += chb.back();
            ++ans2;
            chb.pop_back();
        }
    }
    cout << ans1 << " " << ans2 << endl;
    return 0;
}

6D

不会做.....题解看不懂....... dfs

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 22;
int n, a, b, ans, h[MAXN], hh[MAXN];


bool dfs(int curpos, int time, int hitpos){
    if(time == 0){
        for(int i = 1; i <= n; ++i)
            if(h[i] >= 0) return false;
        return true;
    }
    if(h[curpos] < 0) return dfs(curpos+1, time, hitpos);
    for(int i = min(n-1, max(hitpos, max(2, curpos))); i <= min(n-1, curpos+1); ++i){
        h[i] -= a; h[i-1] -= b; h[i+1] -= b;
        if(dfs(curpos, time-1, i)){
            hh[time] = i;
            return true;
        }
        h[i] += a; h[i-1] += b; h[i+1] += b;
    }
    return false;
}


int main(){
    cin >> n >> a >> b;
    for(int i = 1; i <= n; ++i) cin >> h[i];
    for(int i = 1; ; ++i){
        memset(hh, 0, sizeof(hh));
        if(dfs(1, i, 2)){
            cout << i << endl;
            for(int j = 1; j <= i; ++j)
                cout << hh[j] << " ";
            return 0;
        }
    }
    return 0;
}

6E

求最大值最小值差不超过给定值的最长子串

ST表预处理区间最大值最小值，加个二分右边界...

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 7;
int n, k;
int data[maxn];
int premin[maxn][20], premax[maxn][20];
struct tpoint{
    int l, r;
    tpoint(int _l, int _r): l(_l), r(_r) {};
};

void init_st(){
    for(int i = 1; i <= n; ++i)
        premin[i][0] = premax[i][0] = data[i];
    int k = floor(log(double(n)) / log(2.0));
    for(int j = 1; j <= k; ++j)
        for(int i = n; i >= 1; --i)
            if(i + (1<<(j-1)) <= n){
                premin[i][j] = min(premin[i][j-1], premin[i+(1<<(j-1))][j-1]);
                premax[i][j] = max(premax[i][j-1], premax[i+(1<<(j-1))][j-1]);
            }
}

int query_st(int l, int r){
    int k = floor(log(double(r-l+1)) / log(2.0));
    return max(premax[l][k], premax[r-(1<<k)+1][k]) - min(premin[l][k], premin[r-(1<<k)+1][k]);
}

int check(int i){
    int l = i, r = n + 1;
    while(l < r - 1){
        int mid = (l + r) >> 1;
        if(query_st(i, mid) <= k) l = mid;
        else r = mid;
    }
    return l;
}

int main(){
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &data[i]);
    init_st();
    int amount = 0;
    vector<tpoint> vp;
    for(int l = 1; l <= n; ++l){
        int r = check(l);
        int num = r - l + 1;
        if(num > amount){
            vp.clear();
            vp.push_back(tpoint(l, r));
            amount = num;
        }
        else if(num == amount) vp.push_back(tpoint(l, r));
    }
    printf("%d %d
", amount, vp.size());
    for(vector<tpoint>::iterator i = vp.begin(); i != vp.end(); ++i)
        printf("%d %d
", (*i).l, (*i).r);
    return 0;
}

7A

水题但是有点意思(因为你还是个渣啊~)

记下每一行每一列多少个黑，凑齐8个答案+1，若答案是16就-8

#include <bits/stdc++.h>
using namespace std;

char cb[10][10];
int cnt[10][10];

int main(){
    ios::sync_with_stdio(false);
    int ans = 0;
    for(int i = 0; i < 8; ++i)
        for(int j = 0; j < 8; ++j){
            cin >> cb[i][j];
            if(cb[i][j] == 'B'){
                ++cnt[i][8];
                ++cnt[8][j];
            }
            if(cnt[i][8] == 8) ++ans;
            if(cnt[8][j] == 8) ++ans;
        }
    if(ans == 16) ans -= 8;
    cout << ans << endl;
    return 0;
}

7B

粗暴地模拟，大概坑都是给我挖的....

#include <bits/stdc++.h>
using namespace std;

const int maxn = 111;
int n, k, mem[maxn];

int main(){
    int counter = 0;
    scanf("%d %d", &n, &k);
    while(n--){
        char s[11];
        int x;
        scanf("%s", s);
        if(s[0] == 'a' || s[0] == 'e'){
            scanf("%d", &x);
            if(s[0] == 'a'){
                bool flag = false;
                int tmp = 0;
                for(int i = 0; i < k; ++i){
                    if(mem[i] == 0) ++tmp;
                    else tmp = 0;
                    if(tmp == x){
                        printf("%d
", ++counter);
                        int index = i;
                        while(tmp--) mem[index--] = counter;
                        flag = true;
                        break;
                    }
                }
                if(!flag) puts("NULL");
            }
            else{
                bool flag = false;
                for(int i = 0; i < k; ++i){
                    if(x == 0) break;
                    if(mem[i] == x){
                        mem[i] = 0;
                        flag = true;
                    }
                }
                if(!flag) puts("ILLEGAL_ERASE_ARGUMENT");
            }
        }
        else{
            for(int i = 0; i < k; ++i){
                if(mem[i] != 0){
                    int j = i;
                    while(j > 0 && mem[j-1] == 0) --j;
                    swap(mem[i], mem[j]);
                }
            }
        }
    }
    return 0;
}

7C

扩展欧几里得解不定方程组

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll ex_gcd(ll a, ll b, ll& x, ll& y){
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    int k = ex_gcd(b, a % b, x, y);
    int t = y;
    y = x - (a / b) * y;
    x = t;
    return k;
}

int main(){
    ios::sync_with_stdio(false);
    ll a, b, c, x, y;
    cin >> a >> b >> c;
    ll k = ex_gcd(a, b, x, y);
    c = -c;
    if(c % k) cout << "-1" << endl;
    else cout << (c / k * x) << " " << (c / k * y) << endl;
    return 0;
}

7D

字符串两个方向hash判回文+dp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 5e6 + 7;
const int mod = 1e9 + 7;
const int tmp = 137731735;
char s[maxn];
ll prefix[maxn], suffix[maxn], dp[maxn];

int main(){
    gets(s);
    for(int i = 0; s[i]; ++i){
        if(s[i] >= '0' && s[i] <= '9')
            s[i] -= '0';
        else if(s[i] >= 'a' && s[i] <= 'z')
            s[i] -= ('a' - 10);
        else
            s[i] -= ('A' - 36);
    }
    ll len = strlen(s), base = 1;
    for(int i = 1; i <= len; ++i){
        prefix[i] = (prefix[i-1] + s[i-1] * base) % mod;
        base = (base * tmp) % mod;
        suffix[i] = (suffix[i-1] * tmp + s[i-1]) % mod;
    }
    ll ans = 0;
    for(int i = 1; i <= len; ++i){
        if(prefix[i] == suffix[i]){
            dp[i] = dp[i>>1] + 1;
            ans += dp[i];
        }
    }
    printf("%I64d
", ans);
    return 0;
}

7E

arhgoiawhengiuehgoaijgioj........

8A

水题，用py3写是因为.....懒......

s = input()
s1 = input()
s2 = input()

forw = backw = False
tmp = s.find(s1)
if ~tmp:
    if ~s[tmp+len(s1):].find(s2):
        forw = True
s = s[::-1]
tmp = s.find(s1)
if ~tmp:
    if ~s[tmp+len(s1):].find(s2):
        backw = True

if forw and backw: print('both')
elif forw: print('forward')
elif backw: print('backward')
else: print('fantasy')

8B

啊啊啊啊啊啊啊手贱啊啊啊啊啊啊啊  真·教育题  谢谢(微笑)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> point;
map<point, int> check;
int cx[] = {1, -1, 0, 0}, cy[] = {0, 0, -1, 1};
int main(){
    string s;
    cin >> s;
    int x = 0, y = 0, xx = 0, yy = 0;
    ++check[point(x, y)];
    bool flag = true;
    for(int i = 0; i < s.size(); ++i){
        if(s[i] == 'L') --x;
        else if(s[i] == 'R') ++x;
        else if(s[i] == 'U') ++y;
        else --y;
        for(int i = 0; i < 4; ++i){
            int xxx = x + cx[i], yyy = y + cy[i];
            if(xxx == xx && yyy == yy) continue;
            if(check[point(xxx, yyy)]){
                flag = false;
                break;
            }
        }
        if(check[point(x, y)]) flag = false;
        if(!flag) break;
        ++check[point(x, y)];
        xx = x; yy = y;
    }
    cout << (flag ? "OK" : "BUG") << endl;
    return 0;
}

8C

状压dp，不需要绝对固定的顺序，所以每次只要取最前面可取的(line34的break)，每次拿一个或两个(最里层for从 j 开始)

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 25;
const int INF = 0x3f3f3f3f;
struct point{
    int x, y;
    point(){}
    point(int xx, int yy): x(xx), y(yy) {}
};
point p[MAXN];
int n, dp[1<<MAXN], pre[1<<MAXN], dis[MAXN][MAXN], ans[555];
inline int calc(point a, point b){
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
int main(){
    cin >> p[0].x >> p[0].y >> n, p[n] = p[0];
    for(int i = 0; i < n; ++i) cin >> p[i].x >> p[i].y;
    for(int i = 0; i <= n; ++i) for(int j = 0; j <= n; ++j) dis[i][j] = calc(p[i], p[j]);
    memset(dp, INF, sizeof(dp));
    dp[0] = 0;
    for(int i = 0; i < (1<<n); ++i)
        if(dp[i] != INF)
            for(int j = 0; j < n; ++j)
                if(!(i>>j&1)){
                    for(int k = j; k < n; ++k)
                        if(!(i>>k&1)){
                            int cur = i | (1<<j) | (1<<k);
                            int tmp = dp[i] + dis[n][j] + dis[j][k] + dis[k][n];
                            if(dp[cur] > tmp){
                                dp[cur] = tmp;
                                pre[cur] = i;
                            }
                        }
                    break;
                }
    cout << dp[(1<<n)-1] << endl << 0;
    int cnt = 0;
    for(int i = (1<<n)-1; i; i = pre[i]){
        int tmp = pre[i] ^ i;
        ans[cnt++] = 0;
        for(int j = 0; j < n; ++j)
            if((1<<j)&tmp)
                ans[cnt++] = j + 1;
    }
    for(int i = cnt-1; ~i; --i) cout << " " << ans[i];
    return 0;
}

8D