A
#include<bits/stdc++.h> using namespace std; int dir[4][2] = {{0, -1}, {0, 1}, { -1, 0}, {1, 0}}; typedef long long ll; int n, s; int h[105]; int m[105]; int num[1000005]; int getans(int h1, int m1, int h2, int m2) { int x1 = (h2 - h1) * 60; int x2 = m2 - m1; return x1 + x2; } int main() { //freopen("out.txt", "w", stdout); cin >> n >> s; for (int i = 1; i <= n; i++) { cin >> h[i] >> m[i]; } if (h[1] * 60 + m[1] >= 1 + s) { cout << 0 << " " << 0 << endl; return 0; } for (int i = 1; i <= n - 1; i++) { int now = getans(h[i], m[i], h[i + 1], m[i + 1]); if (now >= s * 2 + 2) { m[i]++; m[i] += s; while(m[i] >= 60) { m[i] -= 60; h[i]++; } cout << h[i] << " " << m[i] << endl; return 0; } } m[n]++; m[n] += s; while (m[n] >= 60) { m[n] -= 60; h[n]++; } cout << h[n] << " " << m[n] << endl; }
B
#include<bits/stdc++.h> using namespace std; int dir[4][2] = {{0, -1}, {0, 1}, { -1, 0}, {1, 0}}; typedef long long ll; int n; ll a, b; ll s[100005]; ll sum = 0; priority_queue<int, vector<int>, less<int> >que; int main() { //freopen("out.txt", "w", stdout); cin >> n >> a >> b; for (int i = 1; i <= n; i++) { cin >> s[i]; sum += s[i]; if (i != 1) { que.push(s[i]); } } int anser = 0; if (s[1]*a >= b * sum) { cout << 0 << endl; return 0; } int flag = 1; while (flag) { sum -= que.top(); que.pop(); anser++; if (s[1]*a >= b * sum) { cout << anser << endl; return 0; } } }
C
注意判定同楼层的情况
#include<bits/stdc++.h> using namespace std; int dir[4][2] = {{0, -1}, {0, 1}, { -1, 0}, {1, 0}}; typedef long long ll; ll l[100005]; ll e[100005]; ll n, m, cl, ce, v; ll x1, y1, x2, y2; ll anser = 0; ll addx, addy; int main() { //freopen("out.txt", "w", stdout); cin >> n >> m >> cl >> ce >> v; for (int i = 1; i <= cl; i++) { cin >> l[i]; } sort(l + 1, l + cl + 1); for (int i = 1; i <= ce; i++) { cin >> e[i]; } sort(e + 1, e + ce + 1); int q; cin >> q; while (q--) { anser = INT_MAX; cin >> x1 >> y1 >> x2 >> y2; if (x1 == x2) { cout << abs(y1 - y2) << endl; continue; } if (y1 > y2) { swap(x1, x2); swap(y1, y2); } addy = abs(x1 - x2); int now = lower_bound(l + 1, l + cl + 1, y1) - l - 1; //cout<<" "<<now<<endl; int lef = max(now - 2, 1); int rig = min(now + 2, (int)cl); //cout << lef << " l1 " << rig << endl; for (int i = lef; i <= rig; i++) { addx = abs(l[i] - y1) + abs(l[i] - y2); anser = min(anser, addx + addy); } now = lower_bound(l + 1, l + cl + 1, y2) - l - 1; lef = max(now - 2, 1); rig = min(now + 2, (int)cl); //cout << lef << " l2 " << rig << endl; for (int i = lef; i <= rig; i++) { addx = abs(l[i] - y1) + abs(l[i] - y2); anser = min(anser, addx + addy); } addy = abs(x1 - x2) / v; if (addy * v < abs(x1 - x2)) { addy++; } now = lower_bound(e + 1, e + ce + 1, y1) - e - 1; lef = max(now - 2, 1); rig = min(now + 2, (int)ce); //cout << lef << " e1 " << rig << endl; for (int i = lef; i <= rig; i++) { addx = abs(e[i] - y1) + abs(e[i] - y2); anser = min(anser, addx + addy); } now = lower_bound(e + 1, e + ce + 1, y2) - e - 1; lef = max(now - 2, 1); rig = min(now + 2, (int)ce); //cout << lef << " e2 " << rig << endl; for (int i = lef; i <= rig; i++) { addx = abs(e[i] - y1) + abs(e[i] - y2); anser = min(anser, addx + addy); } cout << anser << endl; } }
D
卡题意 总共有两个任务 分别需要X1 X2的资源
你有N个服务器 每个服务器上只能运行一个任务 但是你可以把任务平分到几个服务器上运行 问你能不能完成这两个任务
肯定先排序 然后这两个任务各所在的服务器肯定是连续的一段
预处理出每个任务分成i份所需要的资源 枚举哪个任务所在服务器是前面的 然后再枚举这个任务分成几份 check第二个任务是否能满足
能满足就输出
#include<bits/stdc++.h> #define maxn 300005 using namespace std; int a[maxn], b[maxn], c[maxn], id[maxn], n; void solve1() { for (int i = n - 1; i > 0; i--) { int p = lower_bound(c + 1, c + n + 1, a[i]) - c; //如果选x1作为前面连续的一部分 分为i份所需要开始的最前位置 int ps = p + i - 1; //分成i份后的末尾部分 if (ps < n && b[n - ps] <= c[ps + 1]) //如果作为前面一部分成立 并且把x2分成n-ps份后所需的资源数小于c[ps+1] 整体成立 { puts("Yes"); printf("%d %d ", i, n - ps); for (int i = p; i <= ps; i++) { printf("%d ", id[i]); } puts(""); for (int i = ps + 1; i <= n; i++) { printf("%d ", id[i]); } exit(0); } } } void solve2() //作用同上 { for (int i = n - 1; i > 0; i--) { int p = lower_bound(c + 1, c + n + 1, b[i]) - c, ps = p + i - 1; if (ps < n && a[n - ps] <= c[ps + 1]) { puts("Yes"); printf("%d %d ", n - ps, i); for (int i = ps + 1; i <= n; i++) { printf("%d ", id[i]); } puts(""); for (int i = p; i <= ps; i++) { printf("%d ", id[i]); } exit(0); } } } bool cmp(const int &A, const int &B) { return c[A] < c[B]; } int main() { int x1, x2; scanf("%d%d%d", &n, &x1, &x2); for (int i = 1; i <= n; i++) { scanf("%d", &c[i]), id[i] = i; } sort(id + 1, id + n + 1, cmp); //相当于sort piar<int,int> sort(c + 1, c + n + 1); for (int i = 1; i <= n; i++) { a[i] = x1 / i + (x1 % i > 0), b[i] = x2 / i + (x2 % i > 0); //a[i] 表示如果x1平均分为i个所需要的资源数 //b[i] 表示如果x2平均分为i个所需要的资源数 } solve1(); solve2(); puts("No"); return 0; }
E