題意：

給出第一天的時候鳥擁有的食物儲量，給出鳥每天要吃的食物量，接下來有m組數據，每一行代表某只鳥在第幾天的時候收獲了多少食物，輸出所有鳥在最后一次收獲食物之后一天的狀況（死了或者剩多少食物）。

思路：

模擬就好，隊友打的，具體的細節看代碼。

代碼：

#include<cstdio> #include<cstring> #include<climits> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<string> #include<queue> #include<map> #include<vector> #include<set> #include<sstream>using namespace std;const int maxn = 100005; struct B {string name;int x, y; }b[maxn]; bool cmp(B a, B b) {if(a.name != b.name) {return a.name < b.name;}if(a.x != b.x) return a.x < b.x; }int main() {int t, n, s, k; // freopen("a.txt","r",stdin);scanf("%d",&t);for(int kase = 1; kase <= t; kase++) {scanf("%d",&n);scanf("%d %d",&s, &k);for(int i = 0; i < n; i++) {cin >> b[i].name >> b[i].x >> b[i].y;}sort(b, b + n, cmp);// for(int i = 0; i < n; i++) { // cout << b[i].name << " " << b[i].x << " " << b[i].y << endl; // }puts("");printf("Case #%d:\n", kase);bool flag = true;string na = b[0].name;int zong = s;int shang = 1;for(int i = 0; i <= n; i++){if(b[i].name != na || i == n) {if(flag == false) {cout << b[i - 1].name << " died." << endl;} else {cout << b[i - 1].name << " " << zong << endl;}flag = true;zong = s;shang = 1;zong = zong - (b[i].x - shang) * k;if(zong < 0) {flag = false;} else {zong += b[i].y;zong -= k;if(zong < 0) {flag = false;}}shang = b[i].x + 1;na = b[i].name;if(zong < 0) {flag = false;}} else {if(i > 0 && b[i].name == b[i - 1].name && b[i].x == b[i - 1].x) {zong += b[i].y;}zong = zong - (b[i].x - shang) * k;if(zong < 0) {flag = false;} else {zong += b[i].y;zong -= k;if(zong < 0) {flag = false;}}shang = b[i].x + 1;}}puts("");}return 0; }

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