BNUOJ 5227 Max Sum

Max Sum

Time Limit: 1000ms

Memory Limit: 32768KB

 

This problem will be judged on HDU. Original ID: 1003
64-bit integer IO format: %I64d      Java class name: Main

 

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

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


解题：dp入门题！弱菜的成长之路啊！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
int dp[100010],num[100010];
int main(){
    int kase,i,index,ans,n,k = 1;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d",&n);
        for(i = 1; i <= n; i++)
            scanf("%d",dp+i);
        num[1] = 0;
        ans = dp[index = 1];
        for(i = 2; i <= n; i++){
            if(dp[i] <= dp[i-1]+dp[i]){
                dp[i] = dp[i-1]+dp[i];
                num[i] = num[i-1]+1;
            }else num[i] = 0;
            if(ans < dp[i]) ans = dp[index = i];
        }
        printf("Case %d:\n",k++);
        printf("%d %d %d\n",ans,index-num[index],index);
        if(kase) putchar('\n');
    }
    return 0;
}