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自然万物都趋向从有序变得无序

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HDU 1003 Max Sum

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Max Sum

Given a sequence a11,a22,a33......ann, your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

思路

状态转移方程dp[i]=max(dp[i-1]+a[i],a[i]);

代码

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#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

int main(){
    int t;
    int cas=0;
    scanf("%d",&t);
    while(t--){

        int n;
        scanf("%d",&n);
        int m=-1111;
        int s=0;
        int sx=1,sy=1,temp=1;
        for(int i=1;i<=n;++i){
            int x;
            scanf("%d",&x);
            s+=x;
            if(s>m){
                m=s;
                sx=temp;
                sy=i;
            }  if(s<0){
            s=0;
            temp=i+1;
            }
        }

        printf("Case %d:\n%d %d %d\n",++cas,m,sx,sy);
         if(t)
            printf("\n");
    }
    return 0;
}