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[1651] Red packet
- 时间限制: 1000 ms 内存限制: 65535 K
- 问题描述
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New Year is coming! Our big boss Wine93 will distribute some “Red Package”, just like Alipay and Wechat.
Wine93 has m yuan, he decides to distribute them to n people and everyone can get some money(0 yuan is not allowed and everyone’s money is an integer), Now k people has gotten money, it’s your turn to get “Red Package”, you want to know, at least how much money to give you, then you can must become the “lucky man”. and the m yuan must be used out.
Noting that if someone’s money is strictly much than others’, than he is “lucky man”.
- 输入
- Input starts with an integer T (T <= 50) denoting the number of test case.
For each test case, three integers n, m, k (1 <= k < n <= 100000, 0< m <= 100000000) will be given.
Next line contains k integers, denoting the money that k people get. You can assume that the k integers’ summation is no more than m.
- 输出
- Ouput the least money that you need to become the “lucky man”, if it is impossible, output “Impossible” (no quote).
- 样例输入
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3 3 5 2 2 1 4 10 2 2 3 4 15 2 3 5
- 样例输出
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Impossible 4 6
思路：
发红包有成为幸运儿的说法即你要比每个人得到的红包都要多，你就是幸运儿。
输入数据：
测试数据个数
人数，红包总额，已经分发的人数
已经分发的人数的红包
思路：
代码注释部分。
如果描述不清晰：
博主参考自：Classmata_Wang Yinggang
代码：
#include<cstdio>
#include<cstdlib>
#define MYDD 1103int MAX(int x,int y) {return x>y? x:y;
}int main() {int t;scanf("%d",&t);while(t--) {int n,m,k;//人数，红包总额，已经分发的人数int max_get,distri_money,money;scanf("%d%d%d",&n,&m,&k);max_get=0;//当前已分发红包的最高金额distri_money=0;//已经分发红包的总额for(int j=0; j<k; j++) {scanf("%d",&money);distri_money+=money;max_get=MAX(max_get,money);//已经得到红包的最大金额}int remaining_money=m-distri_money;//剩下的金额if(remaining_money<=max_get) {puts("Impossible");//剩下的红包不大于已经得到红包的最大金额continue;}int remaining_people=n-k-1;//没发到红包的人数(不包含幸运儿)if(remaining_people==0) {//此时没有剩下没发的红包printf("%d\n",remaining_money);}int left=max_get+1;//二分求解满足条件int right=remaining_money;while(left<=right) {int middle=(left+right)/2;int k=remaining_money-middle-remaining_people+1;if(k<middle)right=middle-1;elseleft=middle+1;}printf("%d\n",left);}return 0;
}
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