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HDU 1003 Max Sum (动规)
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141547 Accepted Submission(s): 32929
Total Submission(s): 141547 Accepted Submission(s): 32929
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
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此题与HDU 1231一样,不过简单一些,记录前标要容易。
此题与HDU 1231一样,不过简单一些,记录前标要容易。
#include <iostream> using namespace std; #define M 100100 int vis[M],dp[M]; int main(int i,int j,int k) { int n,last,t,first,temp; cin>>t; for(k=1;k<=t;k++) { cin>>n; memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++) {cin>>vis[i];dp[i]=vis[i];} int sum=0,maxSum=vis[1];first=last=temp=1; for(i = 1; i <= n; i++) { sum += vis[i]; //i是从1开始的,所以先加再判断。 if(sum > maxSum){ maxSum = sum; first = temp; last = i; } if(sum < 0){ //这里用sum表示vis[first]->vis[i]的和,所以要清零。 sum = 0; temp = i+1; } } printf("Case %d:\n",k); printf("%d %d %d\n",maxSum,first,last); if(k < t) printf("\n"); 好吧,这里就让我格式错误了几次。。。明明Case 1:和Case 2:数据中间有空行,但是完成一次没空行。。。 } return 0; }
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