LA UVaLive 6855 Banks (水题，暴力)

题意：给定 n 个数，让你求最少经过几次操作，把所有的数变成非负数，操作只有一种，变一个负数变成相反数，但是要把左右两边的数加上这个数。

析：由于看他们AC了，时间这么短，就暴力了一下，就AC了。。。。。并不明白

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std ;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];

int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 0; i < n; ++i)   scanf("%d", &a[i]);
        int ans = 0;
        while(true){
            int t = ans;
            for(int i = 0; i < n; ++i)
                if(a[i] < 0){
                    a[i] = -a[i];  a[(i+1)%n] -= a[i];  a[(i-1+n)%n] -= a[i];
                    ++ans;
                }
            if(ans == t)  break;
        }
        printf("%d\n", ans);
    }
    return 0;
}