NEU OJ 1651 Great number

循环节是2000000016

字符串读入，用一下高精度对2000000016取个模，用一下快速幂就可以算出答案了。

#include <cstdio>
#include <iostream>
#include<cstring>
using namespace std;
const long long MOD = 1e9+;
long long mod1(char *a1,int b)
{

    long long  a[] = {};
    long long c[] = {};

    long long  i, k, d;
    k = strlen(a1);
    for(i = ; i < k; i++)  a[i] = a1[k - i - ] - '';
    d = ;
    for(i = k - ; i >=  ; i--)
    {
        d = d *  + a[i];
        c[i] = d / b;
        d = d % b;
    }
    while(c[k - ] ==  && k > )  k--;
    return d;
}
struct matrix
{
    long long m[][];
}ans, base;

matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = ; i < ; ++i)
    {
        for(int j = ; j < ; ++j)
        {
            tmp.m[i][j] = ;
            for(int k = ; k < ; ++k)
                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]% MOD) % MOD;
        }
    }
    return tmp;
}
int fast_mod(int n)  // 求矩阵 base 的  n 次幂
{
    base.m[][] = base.m[][] = base.m[][] = ;
    base.m[][] = ;
    ans.m[][] = ans.m[][] = ;  // ans 初始化为单位矩阵
    ans.m[][] = ans.m[][] = ;
    while(n)
    {
        if(n & )
        {
            ans = multi(ans, base);
        }
        base = multi(base, base);
        n >>= ;
    }
    return ans.m[][];
}
char SS[];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",SS);
        //printf("%d\n",mod1(SS,2000000016));
        printf("%lld\n", fast_mod(mod1(SS,))%MOD);
    }
    return ;
}