[CFGym101061G] Repeat it（逆元）

时间：2023-03-09 21:24:22
[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 ;

 }

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