BOJ 262 Channel Coding

Description
Assume an integer sequence contains N elements whose value could only be one of {0, 1, -1}. There may exist a positive integer D which can make the sum between i-th element and (i+D)-th element be zero, where i is a certain integer between 1 and N-D. Your task is to find out the maximal D.

Input
First line contains a integer T indicate the number of cases. (T<=50)
In each test case, the first line is N descript above. (N<=50000)
Then follows N integers the sequence contains.

Output
Each case output one integer D in a line. If there is no such a D, just output -1.

Sample Input

2
3
1 0 -1
1
1

Sample Output

2
-1

很简单的hash

#include<stdio.h> #include<stdlib.h>; #include<string.h> #define inf 100001 int Max(int a,int b){ return a>b?a:b; } int Min(int a,int b){ return a>b?b:a; } int max[100100],min[100100],res; int main(){ int i,j,n,T,t,tem; scanf("%d",&T); for(t=1;t<=T;t++){ scanf("%d",&n); int sum=0; res=-1; for(i=0;i<=2*n;i++){ max[i]=0; min[i]=inf; } max[0]=0,min[0]=0; for(i=1;i<=n;i++){ scanf("%d",&tem); sum+=tem; if(sum<0){ min[n-sum]=Min(min[n-sum],i); max[n-sum]=Max(max[n-sum],i); res=Max(res,max[n-sum]-min[n-sum]); } else{ min[sum]=Min(min[sum],i); max[sum]=Max(max[sum],i); res=Max(res,max[sum]-min[sum]); } } if(res>1) printf("%d\n",res-1); else printf("-1\n"); } return 0; }