Hdu 4223 Dynamic Programming?
Dynamic Programming?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2188 Accepted Submission(s): 1005
Problem Description
Dynamic Programming, short for DP, is the favorite of iSea. It is a method for solving complex problems by breaking them down into simpler sub-problems. It is applicable to problems exhibiting the properties of overlapping sub-problems which are only slightly smaller and optimal substructure.
Ok, here is the problem. Given an array with N integers, find a continuous subsequence whose sum’s absolute value is the smallest. Very typical DP problem, right?
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case includes an integer N. Then a line with N integers Ai follows.
Technical Specification
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. -100 000 <= Ai <= 100 000
Output
For each test case, output the case number first, then the smallest absolute value of sum.
Sample Input
2
2
1 -1
4
1 2 1 -2
Sample Output
Case 1: 0
Case 2: 1
#include"stdio.h" #include"string.h" #define N 1101 int abs(int a) { if(a>0)return a; return -a; } int main() { int T; int n; int A[N]; int dp[N]; int i,j,t; int ans,cnt; scanf("%d",&T); cnt=1; while(T--) { scanf("%d",&n); scanf("%d",&A[0]); ans=dp[0]=abs(A[0]); for(i=1;i<n;i++) { scanf("%d",&A[i]); dp[i]=abs(A[i]); t=A[i]; //因为要求是连续的,所以得从i-1开始 for(j=i-1;j>=0;j--) { t+=A[j]; if(abs(t)<dp[i]) dp[i]=abs(t); } if(dp[i]<ans)ans=dp[i]; } printf("Case %d: %d\n",cnt++,ans); } return 0; }