VJ 【规律题】

/*DescriptionSereja 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.InputThe 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).OutputFor each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.Sample InputInputzyxxxxxxyyz55 51 31 111 43 6OutputYESYESNOYESNOHintIn 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.*/ #include<stdio.h>#include<stdlib.h>#include<math.h> #include<string.h>char s[100010];int c[100010], a[100010], b[100010];int main(){int i, j, k, n;while(scanf("%s", s) != EOF){scanf("%d", &n);int len = strlen(s);int x = 0, y = 0, z = 0;for(i = 0; i < len; ++i){if(s[i] == 'x')x++;if(s[i] == 'y')y++;if(s[i] == 'z')z++;a[i] = x;b[i] = y;c[i] = z;}for(i = 0;  i < n; ++i){int l, r;scanf("%d%d", &l, &r);if(r-l+1 < 3)printf("YES\n");else{x = a[r-1] - a[l-2];y = b[r-1] - b[l-2];z = c[r-1] - c[l-2];if(abs(x-y) <= 1 && abs(x-z) <= 1 && abs(y-z) <= 1)printf("YES\n");elseprintf("NO\n");}} }return 0;} //题意；给出一个字符串 ，该字符串可以任意变换，当该字符串存在任意三个连续的字符都满足 "zyx", "xzy", "yxz" 时输出YES //思路：根据题意可以推导出只要xyz字符的个数两两相差的个数不超过一个 就可以得到符合题意的字符串