动态规划: HDU1003Max Sum
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 251171 Accepted Submission(s): 59503
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
Problem : 1003 ( Max Sum ) Judge Status : Accepted
RunId : 21239601 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
RunId : 21239601 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int INF = -(1<<30);int main(){ int T,n,cnt = 0; scanf("%d",&T); while(T--){ int ans ,start=1,tail,tmp = 0, x,ts; scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&x); if(i==1){ ans = tmp = x; start = tail = ts = 1; } else{ if(x > x + tmp){ tmp = x; ts = i; } else tmp = tmp + x; } if(tmp > ans ){ ans = tmp; start = ts,tail = i; } } printf("Case %d:\n%d %d %d\n",++cnt,ans,start,tail); if(T) printf("\n"); }}
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