HDOJ 1003-Max Sum【DP】

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 197162    Accepted Submission(s): 46049

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

解题思路：

dp用于储存累加的和，并和max的值比较如果dp都已经小于零则dp的下一位就不用累加前面的项从下一位开始。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int map[100000];int dp[100000];int main(){int T;int ans=0;scanf("%d",&T);while(T--){ans++;int N;int i,j;scanf("%d",&N);for(i=0;i<N;i++){scanf("%d",&map[i]);}dp[0]=map[0];int start=0,endd=0,maxx=-1001;int frist=0,second=0;for(i=0;i<N;i++){if(dp[i-1]>=0){dp[i]=dp[i-1]+map[i];endd=i;}else{dp[i]=map[i];start=endd=i;}if(maxx<=dp[i]){maxx=dp[i];frist=start;second=endd;}}printf("Case %d:\n",ans);printf("%d %d %d\n",maxx,frist+1,second+1);if(T!=0){printf("\n");}} return 0;}