Topic link:
http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 282195 Accepted Submission(s): 67034
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
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
Author
Ignatius.L
analyze:
The data on the subject is very watery.
Input 6 2 7 -9 5 4 3, the answer should be 12 1 6 The result 12 3 6 can pass! ! ! ! ! ! !
ac code
#include<bits/stdc++.h> using namespace std; int main() { int t; scanf("%d",&t); int T=1; while(T<=t) { int n; scanf("%d",&n); int a[n+1]; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } int dp[n+1],k1=1,k2=1; dp[1]=a[1]; for(int i=2;i<=n;i++) { int x=dp[i-1]; if(x<0) { x=0; } dp[i]=x+a[i]; } int maxvalue=dp[1]; for(int i=2;i<=n;i++) { if(maxvalue<dp[i]) { maxvalue=dp[i]; k2=i; } } for(int i=k2,x=0;x!=maxvalue;--i) { x=x+a[i]; k1=i; } int s=0; for(int i=k2-1;i>=1;i--) { s=s+a[i]; if(s==0) k1=i; } printf("Case %d:\n",T); printf("%d %d %d\n",maxvalue,k1,k2); if(T<t) printf("\n"); T++; } return 0; }