HDU 1003 Max Sum(DP,水题)
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Max Sum
题目链接
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 253355 Accepted Submission(s): 60124
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
简单的DP
题目要求的是连续的最大的和,注意是连续。
知道题意就好办了,可以推出公式 a[i]=max(a[i-1]+b[i],b[i]);
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;int main(){ int a[100010]; int b[100010]; int t; scanf("%d",&t); int num=1; while(t--) { int k; scanf("%d",&k); int maxx=-1001,tt; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); for(int i=1;i<=k;i++) { scanf("%d",&b[i]); a[i]=max(a[i-1]+b[i],b[i]); if(maxx<a[i]) maxx=a[i],tt=i;//记录和的最大值和最后一个的坐标 } int sum=0,ttt; for(int j=tt;j>0;j--) { sum+=b[j]; if(sum==maxx) ttt=j; } printf("Case %d:\n",num++); printf("%d %d %d\n",maxx,ttt,tt); if(t) printf("\n"); } return 0;}
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