回文串的hash解法

来源：互联网发布：java并发库视频教程编辑：程序博客网时间：2024/04/29 19:43
#include <cstring>#include <cstdlib>#include <cstdio>#include <iostream>#include<algorithm>#define maxn 1000010#define inf 31using namespace std;char s[maxn];int  N;unsigned long long bit[maxn],f[maxn],t[maxn];void init(void) {    bit[0] = 1;    for (int i = 1; i <=maxn-10;i++) {        bit[i] = (unsigned long long)bit[i-1]*inf;    }}int low(int x){    return x & -x;}void add(int x, unsigned long long c[], unsigned long long val) {    for (int i = x; i <= N; i += low(i)) {        c[i] += val;    }}unsigned long long sum(int x, unsigned long long c[]) {    unsigned long long ret = 0;    for (int i = x; i > 0; i -= low(i)) {        ret += c[i];    }    return ret;}int panduan(int a, int b) {    int x = a-1, y = N-b, z;    z = max(x, y);    unsigned long long ll = sum(b, f) - sum(a-1, f);    unsigned long long rr = sum(N-a+1, t) - sum(N-b, t);    ll *= bit[z-x];    rr *= bit[z-y];    return ll == rr;}void merg(int x, int val) {    add(x, f, (val-s[x])*bit[x-1]);    add(N+1-x, t, (val-s[x])*bit[N-x]);    s[x] = val;}int main() {    char cc[5], ch[5];    int a, b,Q;    init();    while (scanf("%s", s+1)!=EOF) {        N = (int)strlen(s+1);        memset(f, 0, sizeof (long long) * (N+1));        memset(t, 0, sizeof (long long) * (N+1));        for (int i = 1, j = N; i <= N; ++i, --j) {            s[i] -= 'a';            add(i, f, s[i]*bit[i-1]);            add(j, t, s[i]*bit[j-1]);        }        scanf("%d", &Q);        for(int i=0;i<Q;i++) {            scanf("%s", cc);            if (cc[0] == 'Q') {                scanf("%d %d", &a, &b);                if(panduan(a, b)){                    printf("yes\n");                }                else{                    printf("no\n");                }            } else {                scanf("%d %s", &a, ch);                merg(a, ch[0]-'a');            }        }        printf("\n");    }    return 0;}
0 0