BNU Eugeny and Array
2013-08-12 08:25
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http://www.bnuoj.com/bnuoj/problem_show.php?pid=29452
题意:给你一个只有1和-1组成的数组,然后给你一个范围,求重新排列后在这个范围内求和能否为0,能为0则输出为1,否则为0这个题只要计算一下范围是否为偶数,偶数个数相加才可能为0,然后统计一下1和-1的个数,看里面最少的个数是否大于等于范围的一半。
AC代码:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int n,m,i,j,t,l,r,p; int a[200010]; int b[2]; while(scanf("%d%d",&n,&m)!=EOF) { memset(b,0,sizeof(b)); for(i = 1; i <= n; i++) { scanf("%d",&a[i]); if(a[i] == 1) { b[1]++; } else { b[0]++; } } t = m; while(t--) { scanf("%d%d",&l,&r); p = r-l+1; if(p%2 == 0) { if(p/2 <= b[1] && p/2 <= b[0]) { printf("1\n"); } else { printf("0\n"); } } else { printf("0\n"); } } } return 0; }
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