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 1000000007

using 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;
}

CodeForces 7D Palindrome Degree