codeforces 1405B（思维）

题意描述

You’re given an array a of n integers, such that a1+a2+⋯+an=0.

In one operation, you can choose two different indices i and j (1≤i,j≤n), decrement ai by one and increment aj by one. If i<j this operation is free, otherwise it costs one coin.

How many coins do you have to spend in order to make all elements equal to 0?

可以花费硬币来执行 a [ i ] − 1 , a [ j ] + 1 a[i]-1,a[j]+1 a[i]−1,a[j]+1的操作，如果 i < j i<j i<j那么操作免费，求使得所有元素变为0的最小花费

思路

题意就是让我们用正数来填补负数，所以我们使用一个变量来记录正数的和，碰到负数时用正数和来填补，最后求下所有负数的和，输出相反数即可。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
ll a[N];
void solve(){
    int n;cin>>n;
    rep(i,1,n+1) cin>>a[i];
    ll sum=0;
    rep(i,1,n+1){
        if(a[i]>0) sum+=a[i];
        else if(a[i]<0){
            if(sum+a[i]>=0){
                sum+=a[i];
                a[i]=0;
            }else{
                a[i]+=sum;
                sum=0;
            }
        }
    }
    ll ans=0;
    rep(i,1,n+1){
        if(a[i]<0) ans+=a[i];
    }
    cout<<-ans<<endl;
}
int main(){
    IOS;
    int t;cin>>t;
    while(t--){
        solve();
    }
    return 0;
}