cf C. Sereja and Algorithm

http://codeforces.com/contest/368/problem/C

从左向右记录从1位置到每一个位置上x,y,z的个数。然后判断在l,r区间内的x,y,z的关系满不满足abs(x-y)<=1&&abs(x-z)<=1&&abs(y-z)<=1，满足输出YES，否则输出NO。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 200000
using namespace std;

char str[maxn];
int hx[maxn],hy[maxn],hz[maxn];
int m;
int l,r;

int main()
{
    while(scanf("%s",str)!=EOF)
    {
        int t1=0,t2=0,t3=0;
        memset(hx,0,sizeof(hx));
        memset(hy,0,sizeof(hy));
        memset(hz,0,sizeof(hz));
        int len=strlen(str);
        for(int i=0; i<len; i++)
        {
            if(str[i]=='x')
            {
                t1++;
                hx[i+1]=t1;
                hy[i+1]=t2;
                hz[i+1]=t3;
            }
            else if(str[i]=='y')
            {
                t2++;
                hx[i+1]=t1;
                hy[i+1]=t2;
                hz[i+1]=t3;
            }
            else if(str[i]=='z')
            {
                t3++;
                hx[i+1]=t1;
                hy[i+1]=t2;
                hz[i+1]=t3;
            }
        }
        scanf("%d",&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&l,&r);
            int len=r-l+1;
            if(len<3)
            {
                printf("YES
");
                continue;
            }
            else
            {
                int x=hx[r]-hx[l-1];
                int y=hy[r]-hy[l-1];
                int z=hz[r]-hz[l-1];
                if(abs(x-y)<=1&&abs(x-z)<=1&&abs(y-z)<=1)
                {
                    printf("YES
");
                }
                else printf("NO
");
            }
        }
    }
    return 0;
}