CodeForces 7D Palindrome Degree

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字符串hash。首先说下需要注意的地方：当对Mod取余时，可能造成本不相同的，取余结束之后相同了。

此时应对多个不同的Mod取余，多次计算只能说降低上述情况的发生。感觉正式比赛中不会有这种题，比较拼RP。

比如此题，Mod = 2^32，可以，Mod = 2^64,WA了。。。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <stack>#include <map>#include <ctime>#include <iomanip>#pragma comment(linker, "/STACK:1024000000");#define EPS (1e-6)#define LL long long#define ULL unsigned long long#define _LL __int64#define INF 0x3f3f3f3f#define Mod 1000000007using namespace std;const int MAXN = 5000010;char s[MAXN];ULL e = 256;ULL pre[MAXN],suf[MAXN];ULL p1[MAXN],s1[MAXN];LL K[MAXN];int main(){    scanf("%s",s+1);    int i,n = strlen(s+1);    for(i = 1,pre[0] = 0;i <= n; ++i)        pre[i] = pre[i-1]*e + s[i];    for(i = 2,suf[1] = s[1];i <= n; ++i)        suf[i] = suf[i-1] + s[i]*e,e *= 256;    e = 256;    for(i = 1,p1[0] = 0;i <= n; ++i)    {        p1[i] = p1[i-1]*e + s[i];        p1[i] %= Mod;    }    for(i = 2,s1[1] = s[1];i <= n; ++i)    {        s1[i] = s1[i-1] + s[i]*e,e *= 256;        e %= Mod,s1[i] %= Mod;    }    K[0] = 0;    for(K[0] = 0,i = 1;i <= n; ++i)    {        if(pre[i] == suf[i] && p1[i] == s1[i])            K[i] = K[i/2] + 1;        else            K[i] = 0;    }    LL ans = 0;    for(i = 1;i <= n; ++i)        ans += K[i];    printf("%I64d\n",ans);    return 0;}

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