Beautiful Numbers

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http://codeforces.com/problemset/problem/300/C

题意：给你三个数a,b,n;求满足由a,b组成的n位的个数，且每个位置上的数之和也是用a,b组成;

解析：由题意设a的个数为x,b的个数为y,那么x+y==n;因此枚举满足条件的x的值，然后对这x个a和y进行排列组合。

满足条件的个数为n!/(x!*y!);直接求解会超时。

因此，对该等式进行求逆元，A×inv( b ) % Mod;inv( b ) = pow( b , Mod - 2 );

带入求解。

// File Name: c.cpp// Author: bo_jwolf// Created Time: 2013年10月08日 星期二 15:11:26#include<vector>#include<list>#include<map>#include<set>#include<deque>#include<stack>#include<bitset>#include<algorithm>#include<functional>#include<numeric>#include<utility>#include<sstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cmath>#include<cstdlib>#include<cstring>#include<ctime>using namespace std;const int maxn = 1000005;long long num[ maxn ];#define Mod 1000000007bool judge( long long n, int x, int y ){while( n ){int temp = n % 10;if( temp != x && temp != y )return 0;n /= 10;}return 1;}long long Pow_mod( long long a, long long b ){long long ans = 1;while( b ){if( b & 1 ){ans = ( ans * a ) % Mod;b--;}a = ( a * a ) % Mod;b >>= 1;}return ans;}int main(){num[ 0 ] = 1;for( int i = 1; i < maxn; ++i )num[ i ] = ( num[ i - 1 ] * i ) % Mod;int a, b, n;long long ans;while( scanf( "%d%d%d", &a, &b, &n ) != EOF ){ans = 0;for( int i = 0; i <= n; ++i ){int j = n - i;if( judge( ( i * a + j * b ), a, b ) ){ans += ( num[ n ] * Pow_mod( num[ i ], Mod - 2 ) )% Mod * ( Pow_mod( num[ n - i ], Mod - 2 ) ) % Mod  ;ans %= Mod;}}printf( "%lld\n", ans );}return 0;}