这题范围不会超long long全用int存就行了
贪心的话，每次把一个任务加入到队列，如果不能在指定时间完成就到前面找a最小的一个任务补偿时间，当一个任务完成时间等于0的时候这个任务就不再放回队列
#include<cstdio>#include<queue>#include<algorithm>#include<cstring>using namespace std;//typedef long long LL;const int maxn = 100005;int n;struct Node{int a,b;int d;Node(int a = 0,int b = 0,int d = 0):a(a),b(b),d(d){};friend bool operator <(Node p,Node q){if(p.a != q.a)return p.a < q.a;elsereturn p.d < q.d;}}node[maxn];bool cmp1(Node p,Node q){return p.d < q.d;}int main(){while(scanf("%d",&n) != EOF){for(int i = 0; i < n; i++){scanf("%d%d%d",&node[i].a,&node[i].b,&node[i].d);}sort(node,node + n,cmp1);priority_queue<Node>q;int sum = 0;double ans = 0.0;for(int i = 0; i < n; i++){q.push(node[i]);sum += node[i].b;if(sum > node[i].d){while(sum != node[i].d){Node now = q.top(); q.pop();int need = sum – node[i].d;//需要补偿的时间if(now.b <= need){ans += 1.0 * now.b / now.a;sum -= now.b;}else{now.b -= need;ans += 1.0 * need / now.a;sum -= need;q.push(now);}}}}printf("%.2f\n",ans);}return 0;}
，那么威尼斯就是一艘轻盈和流动的舟船，
