题意:十进制的每一位仅由a和b组成的数是“X数”,求长度为n,各数位上的数的和是X数的X数的个数

思路:由于总的位数为n,每一位只能是a或b,令a有p个,则b有(n-p)个,如果 a*p+b*(n-p) 为X数,则这种情况的答案就是C(n,p),将所有情况累加起来即可。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; #define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull; //#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0);
const int INF = 1e9 + ;
const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ template<int mod>
struct ModInt {
const static int MD = mod;
int x;
ModInt(ll x = ): x(x % MD) {}
int get() { return x; } ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
ModInt operator / (const ModInt &that) const { return *this * that.inverse(); } ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
ModInt operator -= (const ModInt &that) { x -= that.x; if (x < ) x += MD; }
ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
ModInt operator /= (const ModInt &that) { *this = *this / that; } ModInt inverse() const {
int a = x, b = MD, u = , v = ;
while(b) {
int t = a / b;
a -= t * b; std::swap(a, b);
u -= t * v; std::swap(u, v);
}
if(u < ) u += MD;
return u;
} };
typedef ModInt<> mint; const int maxn = 1e6 + ;
bool yes[ * maxn];
int a, b, n;
mint fac[maxn], facinv[maxn]; void pre_init() {
fillchar(yes, );
for (int i = ; i < ( << ); i ++) {
int buf = ;
for (int j = ; j < ; j ++) {
if (( << j) & i) buf = buf * + b;
else buf = buf * + a;
yes[buf] = true;
}
yes[buf] = true;
}
fac[] = facinv[] = ;
for (int i = ; i <= n; i ++) {
fac[i] = fac[i - ] * i;
facinv[i] = facinv[i - ] / i;
}
} int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
while (cin >> a >> b >> n) {
pre_init();
mint ans = ;
for (int i = ; i <= n; i ++) {
if (yes[b * n + (a - b) * i]) {
ans += fac[n] * facinv[i] * facinv[n - i];
}
}
cout << ans.get() << endl;
}
return ;
}
05-04 10:47