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... qk. The algorithm consists of two steps:
- 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.
- 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... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
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 li, ri (1 ≤ li ≤ ri ≤ n).
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
zyxxxxxxyyz 5 5 5 1 3 1 11 1 4 3 6
YES YES NO YES NO
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
题意：一开始真的被题意纠结了好久，然后按自己的理解去写居然就过了，大概就是，在给出的区间内的字符串，如果能通过变换找出任何一组不满足"zyx", "xzy", "yxz"的子串则输出NO，否则YES，小于3的直接输出YES
思路：因为会超时，所以一开始就用哈希记录相应位置x,y,z的数量,然后再每次属入区间时根据哈希计算即可
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char str[100005];
int hash_x[100005],hash_y[100005],hash_z[100005];
int len;
int abs(int a)
{
return a<0?-a:a;
}
int main()
{
int i,j,x,y,z;
while(~scanf("%s",str))
{
len = strlen(str);
x = 0;
y = 0;
z = 0;
memset(hash_x,0,sizeof(hash_x));
memset(hash_y,0,sizeof(hash_y));
memset(hash_z,0,sizeof(hash_z));
for(i = 0;i<len;i++)
{
if(str[i] == 'x')
x++;
else if(str[i] == 'y')
y++;
else
z++;
hash_x[i] = x;
hash_y[i] = y;
hash_z[i] = z;
}
int m,l,r;
scanf("%d",&m);
while(m--)
{
scanf("%d%d",&l,&r);
if(r-l+1<3)
printf("YES\n");
else
{
x = hash_x[r-1]-hash_x[l-2];
y = hash_y[r-1]-hash_y[l-2];
z = hash_z[r-1]-hash_z[l-2];
if(abs(x-y) <= 1 && abs(x-z)<=1 && abs(y-z)<=1)
printf("YES\n");
else
printf("NO\n");
}
}
}
return 0;
}