杭电1003---Max Sum

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 3076   Accepted Submission(s) : 552

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], 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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

Author

Ignatius.L
分析：这是一道动态规划题，刚开始做的时候想了好久，第一次接触感觉真的好难，提交了好多次，，，终于过了

#include<iostream>#include<stdio.h>using namespace std;int main(){    int test;    int n;    int a[100050];    cin>>test;    int k=1;    while(test--)    {        cin>>n;        for(int i=0;i<n;i++)            cin>>a[i];        int max=a[0],now=a[0];        int p1=0,p2=0;       int x=0;        for(int i=1;i<n;i++)        {            if(now+a[i]<a[i])               {                x=i;                now=a[i];            }            else                now=now+a[i];            if(max<now)            {                p1=x;                max=now;                p2=i;            }        }        printf("Case %d:\n",k++);        printf("%d %d %d\n",max,p1+1,p2+1);        if(test)            printf("\n");    }    return 0;}