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
题意:给你一段数组求你的连续的绝对值最小的序列并输出
思路:暴力枚举每一段
代码:
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define inf 0x3f3f3f3f3f3f
using namespace std;
long long a[1000009],dp[1000009];
int main()
{
long long t,n,min1;
scanf("%lld",&t);
long long g=1;
while(t--)
{
min1 = inf;
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
scanf("%lld",&n);
for(long long i=1;i<=n;++i)
{
scanf("%lld",&a[i]);
a[i] = a[i-1]+a[i];//存前几项和
}
for(long long i=1;i<=n;++i)
{
dp[i] = inf;
for(long long j=0;j<i;++j)
{
dp[i] = min(dp[i],abs(a[i]-a[j]));//枚举每一段
}
min1 = min(dp[i],min1);
}
printf("Case %lld: %lld\n",g,min1);//输出
g++;
}
return 0;
}