2016ACM/ICPC亚洲区大连站-重现赛

题目链接：http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C1%AC%D5%BE-%D6%D8%CF%D6%C8%FC%A3%A8%B8%D0%D0%BB%B4%F3%C1%AC%BA%A3%CA%C2%B4%F3%D1%A7%A3%A9&source=1&searchmode=source

A.染色乱搞。

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

typedef struct Edge {
    int to, next;
}Edge;

const int maxn = 1010;
int n, m, x, y;
int know[maxn], vis[maxn], color[maxn];
int ecnt, head[maxn];
Edge e[maxn*maxn];

void init() {
    ecnt = 0;
    memset(head, -1, sizeof(head));
    memset(know, 0, sizeof(know));
    memset(vis, 0, sizeof(vis));
    memset(color, 0, sizeof(color));
}

void adde(int u, int v) {
    e[ecnt].to = v, e[ecnt].next = head[u];
    head[u] = ecnt++;
}

bool dfs(int u) {
    for(int i = head[u]; ~i; i=e[i].next) {
        int v = e[i].to;
        if(color[v] == color[u]) return 0;
        if(vis[v]) continue;
        vis[v] = 1;
        color[v] = 3 - color[u];
        if(!dfs(v)) return 0;
    }
    return 1;
}

int main() {
    // freopen("in", "r", stdin);
    int u, v;
    while(~scanf("%d%d%d%d",&n,&m,&x,&y)) {
        init();
        for(int i = 0; i < m; i++) {
            scanf("%d%d",&u,&v);
            adde(u, v); adde(v, u);
            know[u] = know[v] = 1;
        }
        for(int i = 0; i < x; i++) {
            scanf("%d", &u);
            color[u] = 1; know[u] = 1;
        }
        for(int i = 0; i < y; i++) {
            scanf("%d", &u);
            color[u] = 2; know[u] = 1;
        }
        bool flag = 0;
        for(int i = 1; i <= n; i++) {
            if(!know[i]) {
                flag = 1;
                break;
            }
        }
        if(flag) {
            puts("NO");
            continue;
        }
        for(int i = 1; i <= n; i++) {
            if(color[i]) {
                vis[i] = 1;
                if(!dfs(i)) {
                    flag = 1;
                    break;
                }
            }
        }
        for(int i = 1; i <= n; i++) {
            if(!color[i]) {
                vis[i] = 1;
                color[i] = 1;
                if(!dfs(i)) {
                    flag = 1;
                    break;
                }
            }
        }
        if(!flag) puts("YES");
        else puts("NO");
    }
    return 0;
}

C.威佐夫博弈+大数，根号5用mathmatica跑出来的，其实可以用二分打表。

import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
import java.math.RoundingMode;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        BigDecimal n, m;
        BigDecimal sqrt5 = new BigDecimal("2.2360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081761507737835");
        while(in.hasNextBigDecimal()) {
            n = in.nextBigDecimal(); m = in.nextBigDecimal();
            if(m.equals(n.max(m))) {
                BigDecimal x = n;
                n = m;
                m = x;
            }
            BigDecimal k = n.subtract(m);
            BigDecimal p = new BigDecimal(1);
            n = p.add(sqrt5).multiply(k);
            n = n.divide(new BigDecimal(2));
            if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
            else System.out.println("1");
        }
    }
}

二分打sqrt(5)

import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
import java.math.RoundingMode;
import java.util.Scanner;

public class Main {
        public static BigDecimal Get(BigDecimal x) {
            BigDecimal lo = new BigDecimal("0.000000000000000000000000000001");
            BigDecimal hi = x;
            for(int i = 0; i < 1000; i++) {
                BigDecimal mid = lo.add(hi).divide(new BigDecimal("2.0"));
                BigDecimal mid2 = mid.multiply(mid);
                if(mid2.max(x).equals(mid2)) hi = mid;
                else lo = mid;
            }
            return hi;
        }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        BigDecimal n, m;
        BigDecimal sqrt5 = Get(new BigDecimal(5));
        while(in.hasNextBigDecimal()) {
            n = in.nextBigDecimal(); m = in.nextBigDecimal();
            if(m.equals(n.max(m))) {
                BigDecimal x = n;
                n = m;
                m = x;
            }
            BigDecimal k = n.subtract(m);
            BigDecimal p = new BigDecimal(1);
            n = p.add(sqrt5).multiply(k);
            n = n.divide(new BigDecimal(2));
            if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
            else System.out.println("1");
        }
    }
}

D.找到规律gcd(a,b)=gcd(x,y)以后，就是解方程了。

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

typedef long long LL;
LL a, b, x, y;

LL gcd(LL x, LL y) {
    return y == 0 ? x : gcd(y, x % y);
}
int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%I64d%I64d",&a,&b)) {
        LL g = gcd(a, b);
        LL k = b * g;
        if(a * a - 4 * k < 0 || (LL)sqrt(a*a-4*k) != sqrt(a*a-4*k)) puts("No Solution");
        else {
            LL delta = (LL)sqrt(a * a - 4 * k);
            printf("%I64d %I64d
", (a-delta)/2, (a+delta)/2);
        }
    }
    return 0;
}

F.考虑让整个数列a最长，那么就是公差为1的等差数列，这时候一定会有不够或者超过x的情况。这时候就把多余的部分从后往前+1补齐。问题就变成在这样一个数列里找到最长的不大于x的个数。由于有取模，所以除法用逆元来做。

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

typedef long long LL;

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    else {
        LL ret = exgcd(b, a%b, x, y);
        LL tmp = x;
        x = y;
        y = tmp - a / b * y;
        return ret;
    }
}

LL inv(LL a, LL m) {
    LL x, y;
    exgcd(a, m, x, y);
    return (x % m + m) % m;
}

const LL mod = (LL)1e9+7;
const int maxn = 60600;
LL f[maxn];
int x;

void init() {
    memset(f, 0, sizeof(f));
    f[0] = 1; f[1] = 1;
    for(int i = 2; i < maxn; i++) {
        f[i] = (f[i-1] * i) % mod;
    }
}

int main() {
    // freopen("in", "r", stdin);
    init();
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &x);
        if(x <= 1) {
            printf("%d
", x);
            continue;
        }
        int lo = 2, hi = maxn;
        int pos = 2;
        while(lo <= hi) {
            int mid = (lo + hi) >> 1;
            LL s = (mid + 2) * (mid - 1) / 2;
            if(s <= x) {
                lo = mid + 1;
                pos = max(pos, mid);
            }
            else hi = mid - 1;
        }
        x -= (pos + 2) * (pos - 1) / 2;
        if(x == pos) printf("%I64d
", (f[pos]*inv(2,mod))%mod*(pos+2)%mod);
        else if(pos == 0) printf("%I64d
", f[pos]);
        else printf("%I64d
", f[pos+1]*inv(pos-x+1,mod)%mod);
    }
    return 0;
}

H.奇数就无所谓，偶数的话先手有优势。

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

int k;

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d", &k)) {
        if(k & 1) puts("0");
        else puts("1");
    }
    return 0;
}

I.余弦定理求出第三条边，高用角的一半的余弦乘一下斜边，面积加起来。

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

const int maxn = 11;
const double pi = 3.141592653;
int n, d;
int th;
double ret;

double f(int th) {
    return d*cos(pi*th/2.0/180.0)*sqrt(2.0*d*d-2.0*d*d*cos(pi*th/180.0))/2.0;
}

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d%d",&n,&d)) {
        ret = 0.0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &th);
            ret += f(th);
        }
        printf("%.3lf
", ret);
    }
    return 0;
}

J.拆成四部分挨个看看是不是97。

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

const int maxn = 110;
int n;
int a;

int f(int x) {
    int cnt = 0;
    int q = 4;
    while(q--) {
        if((x % 256) == 97) cnt++;
            x >>= 8;
    }
    return cnt;
}

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d", &n)) {
        int ret = 0;
        for(int i = 1;  i <= n; i++) {
            scanf("%d", &a);
            ret += f(a);
        }
        printf("%d
", ret);
    }
    return 0;
}