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 
     
      #include 
      
       #include
       
        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;}