这道题是武大校赛的题目，和我们OJ以前月赛时的题目一样，只不过我们OJ当时还让求绝对值最大值了，这道题目只让求绝对值最小了。直接用以前的代码就可以ac。就这道题目来说，因为让求绝对值最小的和，所以我们可以先用sum[i]数组保留原数组的前i项和，再对sum数组从小到大排序。这样，最小值就在sum[i]-sum[i-1]和sum[i-1]之间，循环比较久可以了。至于求最大值，排好序后，最大值只可能是sum[n],sum[1],sum[n]-sum[1]中的一个，比较即可。题目：
Dynamic Programming?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 24 Accepted Submission(s): 12
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?
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.
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
ac代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <climits>
#include <string.h>
using namespace std;
const int N=1010;
int a[N],sum[N];
int main(){
//freopen("1.txt","r",stdin);
int ncase,n,kk=0;
scanf("%d",&ncase);
while(ncase--){
scanf("%d",&n);
memset(sum,0,sizeof(sum));
memset(a,0,sizeof(a));
for(int i=0;i<n;++i){
scanf("%d",&a[i]);
}
sum[0]=a[0];
for(int i=1;i<n;++i)
sum[i]=sum[i-1]+a[i];
sort(sum,sum+n);
int mmax=0;
if(fabs((double)(sum[n-1]))>mmax)
mmax=fabs((double)(sum[n-1]));
if(fabs((double)(sum[0]))>mmax)
mmax=fabs((double)(sum[0]));
if(fabs((double)(sum[n-1]-sum[0]))>mmax)
mmax=fabs((double)(sum[n-1]-sum[0]));
int mmin=fabs((double)(sum[0]));
for(int i=0;i<n-1;++i){
if(fabs((double)(sum[i+1]-sum[i]))<mmin)
mmin=fabs((double)(sum[i+1]-sum[i]));
if(fabs((double)(sum[i]))<mmin)
mmin=fabs((double)(sum[i]));
}
//printf("%d %d\n",mmax,mmin);
printf("Case %d: %d\n",++kk,mmin);
}
return 0;
}