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+7;
long long mod1(char *a1,int b)
{

    long long  a[5000] = {0};
    long long c[5000] = {0};

    long long  i, k, d;
    k = strlen(a1);
    for(i = 0; i < k; i++)  a[i] = a1[k - i - 1] - ‘0‘;
    d = 0;
    for(i = k - 1; i >= 0 ; i--)
    {
        d = d * 10 + a[i];
        c[i] = d / b;
        d = d % b;
    }
    while(c[k - 1] == 0 && k > 1)  k--;
    return d;
}
struct matrix
{
    long long m[2][2];
}ans, base;

matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = 0; i < 2; ++i)
    {
        for(int j = 0; j < 2; ++j)
        {
            tmp.m[i][j] = 0;
            for(int k = 0; k < 2; ++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[0][0] = base.m[0][1] = base.m[1][0] = 1;
    base.m[1][1] = 0;
    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵
    ans.m[0][1] = ans.m[1][0] = 0;
    while(n)
    {
        if(n & 1)
        {
            ans = multi(ans, base);
        }
        base = multi(base, base);
        n >>= 1;
    }
    return ans.m[0][1];
}
char SS[1000];
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,2000000016))%MOD);
    }
    return 0;
}