Codeforce-CodeCraft-20 (Div. 2)-A. Grade Allocation

时间:2022-07-28
本文章向大家介绍Codeforce-CodeCraft-20 (Div. 2)-A. Grade Allocation,主要内容包括其使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

n students are taking an exam. The highest possible score at this exam is m. Let ai be the score of the i-th student. You have access to the school database which stores the results of all students.

You can change each student’s score as long as the following conditions are satisfied:

All scores are integers 0≤ai≤m The average score of the class doesn’t change. You are student 1 and you would like to maximize your own score.

Find the highest possible score you can assign to yourself such that all conditions are satisfied.

Input Each test contains multiple test cases.

The first line contains the number of test cases t (1≤t≤200). The description of the test cases follows.

The first line of each test case contains two integers n and m (1≤n≤103, 1≤m≤105) — the number of students and the highest possible score respectively.

The second line of each testcase contains n integers a1,a2,…,an (0≤ai≤m) — scores of the students.

Output For each testcase, output one integer — the highest possible score you can assign to yourself such that both conditions are satisfied._

Example inputCopy 2 4 10 1 2 3 4 4 5 1 2 3 4 outputCopy 10 5 Note In the first case, a=[1,2,3,4], with average of 2.5. You can change array a to [10,0,0,0]. Average remains 2.5, and all conditions are satisfied.

In the second case, 0≤ai≤5. You can change a to [5,1,1,3]. You cannot increase a1 further as it will violate condition 0≤ai≤m. 水体没得说

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    t f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}
 
#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 50050
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
int a[N];
int main()
{
    int t,n;
    long long m;
    read(t);
    while(t--)
    {
        long long ans=0;
        read(n),read(m);
        rep(0,n,i)
        {
            read(a[i]);
            ans+=a[i];
        }
        cout<<min(ans,m)<<endl;
    }
}