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;
}
