Codeforces 490C Hacking Cypher(暴力)

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题目链接：Codeforces 490C Hacking Cypher

分成的两个数字不能有前导0，用复杂度为o(n)的递推方法处理出每个前缀模A，后缀模B的值，找到位置对应前后缀模A、B所得值均为0。

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;
const int maxn = 1e6 + 5;int N, A, B, l[maxn], r[maxn], t[maxn];
char s[maxn];int solve () {t[0] = 1;for (int i = 1; i <= N; i++)t[i] = t[i-1] * 10 % B;for (int i = 0; i < N; i++)l[i] = (l[i-1] * 10 + (s[i] - '0')) % A;int tmp = 0;for (int i = N - 1; i; i--) {tmp = (tmp + (s[i] - '0') * t[N-i-1]) % B;if (s[i] == '0')continue;if (tmp == 0 && l[i-1] == 0)return i;}return 0;
}int main () {scanf("%s%d%d", s, &A, &B);N = strlen(s);int ans = solve();if (ans) {printf("YES\n");for (int i = 0; i < ans; i++)printf("%c", s[i]);printf("\n");for (int i = ans; i < N; i++)printf("%c", s[i]);printf("\n");} elseprintf("NO\n");return 0;
}

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