[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。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <cassert>
 24 #include <cstdio>
 25 #include <bitset>
 26 #include <vector>
 27 #include <deque>
 28 #include <queue>
 29 #include <stack>
 30 #include <ctime>
 31 #include <set>
 32 #include <map>
 33 #include <cmath>
 34 //#include <unordered_map>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%I64d", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70 
 71 const LL mod = 1000000007;
 72 LL n, m;
 73 LL len;
 74 
 75 LL quickmul(LL x, LL q) {
 76     LL ret = 1;
 77     while(q) {
 78         if(q & 1) ret = (ret * x) % mod;
 79         x = (x * x) % mod;
 80         q >>= 1;
 81     }
 82     return ret;
 83 }
 84 
 85 LL exgcd(LL a, LL b, LL &x, LL &y) {
 86     if(b == 0) {
 87         x = 1;
 88         y = 0;
 89         return a;
 90     }
 91     else {
 92         LL ret = exgcd(b, a%b, x, y);
 93         LL tmp = x;
 94         x = y;
 95         y = tmp - a / b * y;
 96         return ret;
 97     }
 98 }
 99 
100 LL mod_inverse(LL a, LL m) {
101     LL x, y;
102     exgcd(a, m, x, y);
103     return (x % m + m) % m;
104 }
105 
106 int main() {
107 // FRead();
108     int T;
109     Rint(T);
110     W(T) {
111         cin >> m >> n;
112         len = 0;
113         LL tmp = n;
114         while(tmp) {
115             len++;
116             tmp /= 10;
117         }
118         LL k = m * len;
119         if(n == 0) len = 1;
120         cout << (n*(quickmul(10,1LL*len*m)-1)%mod*mod_inverse(quickmul(10,len)-1,mod))%mod << endl;
121     }
122     RT 0;
123 }

 

posted @ 2016-08-16 15:07  Kirai  阅读(262)  评论(0)  编辑  收藏  举报