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1050.cpp
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97 lines (85 loc) · 2.11 KB
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/*************************************************************************
> File Name: 1050.cpp
> Author: dulun
> Mail: [email protected]
> Created Time: 2016年03月03日 星期四 15时31分07秒
************************************************************************/
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
using namespace std;
/*“循环数组得最大子段和”问题，可以把环从任意位置断开，
然后求出最优解 = max(普通的最大子段和， 总和 – 普通的“最小子段和”)*/
const int N = 50086;
int a[N];
int main()
{
int n;
LL ans = 0, t = 0;
scanf("%d", &n);
LL sum = 0;
int flag = 0;
int f1 = 0;
for(int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(a[i] > 0) f1 = 1;
}
/*
if(f1 == 0) //全部是负数，输出最大的负数
{
int max = a[0];
for(int i = 0; i < n; i++)
{
if(a[i] > max) max = a[i];
}
cout<<max;
return 0;
}
*/
/*应该是错误思想
for(int i = 0; i < n; i++)
{
if(a[i] < 0) flag = i+1; //只要a[i] < 0 就有可能从a[i] 断开，进行循环
if(t + a[i] < 0 ) { t = 0;}
else t += a[i];
if(t > ans) ans = t;
if(i == n-1 )
{
for(int j = 0; j < flag; j++)
{
if(t + a[j] < 0 ) { t = 0; break;}
else t += a[j];
if(t > ans) ans = t;
}
}
}
cout<<ans;
*/
//求最小子段
LL min = a[0];
t = 0;
for(int i = 0; i < n; i++)
{
if(t + a[i] > 0) t = 0;
else t += a[i];
if(min > t) min = t;
}
LL ans1 = sum - min; //求数组和与最小子段相减
//求最大子段
t = 0;
LL ans2 = 0;
for(int i = 0; i < n; i++)
{
if(t + a[i] < 0 ) { t = 0;}
else t += a[i];
if(t > ans2) ans2 = t;
}
ans = max(ans1, ans2);
cout<<ans;
return 0;
}