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