CF-Sereja and Algorithm

Sereja and Algorithm

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q₁q₂... q_k. The algorithm consists of two steps:

1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2.
2. Rearrange the letters of the found subsequence randomly and go to step 1.

Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.

Sereja wants to test his algorithm. For that, he has string s = s₁s₂... s_n, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring s_lis_li + 1... s_ri(1 ≤ l_i ≤ r_i ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (l_i, r_i) determine if the algorithm works correctly on this test or not.

Input

The first line contains non-empty string s, its length (n) doesn't exceed 10⁵. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.

The second line contains integer m(1 ≤ m ≤ 10⁵) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers l_i, r_i(1 ≤ l_i ≤ r_i ≤ n).

Output

For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.

Sample Input

Input

zyxxxxxxyyz55 51 31 111 43 6

Output

YESYESNOYESNO

解体思路:这道题的题意太难理解了.意思是说,给你一串字符,m次操作,每一次取一个[li,ri]的连续子序列，如果子序列长度小于3，输出YES，如果对于该子序列无论怎样排列，都可以找出3个连续字符，既不是"zyx",也不是 "xzy",也不是 "yxz".那么输出NO，其他情况输出YES。那么我们可以从中找到规律，分别计算[li,ri]区间内（长度大于等于3），x,y,z，的个数，最大个数-最小个数〉1，则输出NO，否则YES。

代码如下：

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;char s[100010];int num[3];int x[100010];int y[100010];int z[100010];int main(){int t,i,l,r,X,Y,Z;while(scanf("%s",s)!=EOF){        int L=strlen(s);        X=Y=Z=0;        for(i=0;i<L;i++){        if(s[i]=='x') {        X++;        }        else if(s[i]=='y'){        Y++;        }        else{        Z++;        }        x[i]=X;y[i]=Y;z[i]=Z;        }scanf("%d",&t);while(t--){scanf("%d%d",&l,&r);    if(r-l+1<3){    printf("YES\n");    continue;    }    if(l==1){     X=x[r-1];      Y=y[r-1];      Z=z[r-1];    }    else{    X=x[r-1]-x[l-2];    Y=y[r-1]-y[l-2];    Z=z[r-1]-z[l-2];    }   int max=-1;   int min=INF;   if(X>max) max=X;   if(Y>max) max=Y;   if(Z>max) max=Z;   if(X<min) min=X;   if(Y<min) min=Y;   if(Z<min) min=Z;   if(max-min>1)printf("NO\n");   else printf("YES\n");}}return 0;}