[CFGym101061G] Repeat it（逆元）

题目链接：http://codeforces.com/gym/101061/problem/G

题意：给一个数字n，让你重复m次，求最后这个数对1e9+7取模的结果。

思路：设数字n长度为k，重复m次即　Σ(i,0->m-1)pow(10, k*i)*n。化简公式得到最终结果为n*(pow(10,k*m)-1)/(pow(10,k)-1)，要取模所以求一发分母的逆元然后乘进去就行了。我是用exgcd求的逆元，群里有巨巨直接用快速幂把逆元搞出来了，暂时还不太明白原理。

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 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 //#include <unordered_map>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const LL mod = ;
 LL n, m;
 LL len;

 LL quickmul(LL x, LL q) {
     LL ret = ;
     while(q) {
         if(q & ) ret = (ret * x) % mod;
         x = (x * x) % mod;
         q >>= ;
     }
     return ret;
 }

 LL exgcd(LL a, LL b, LL &x, LL &y) {
     if(b == ) {
         x = ;
         y = ;
         return a;
     }
     else {
         LL ret = exgcd(b, a%b, x, y);
         LL tmp = x;
         x = y;
         y = tmp - a / b * y;
         return ret;
     }
 }

 LL mod_inverse(LL a, LL m) {
     LL x, y;
     exgcd(a, m, x, y);
     return (x % m + m) % m;
 }

 int main() {
 // FRead();
     int T;
     Rint(T);
     W(T) {
         cin >> m >> n;
         len = ;
         LL tmp = n;
         while(tmp) {
             len++;
             tmp /= ;
         }
         LL k = m * len;
         if(n == ) len = ;
         cout << (n*(quickmul(,1LL*len*m)-)%mod*mod_inverse(quickmul(,len)-,mod))%mod << endl;
     }
     RT ;
 }