莫的难题

题目链接

题目大意

给你五个数，每个数字可以重复选，让你找出第C(n,m) % 1e9+7大的数。

题解

可以先写出C(n,m)的表，
于是题就变成了求由这些数组合成的第k大的数。
怎么求？
把k变成五进制~~（哇妙啊 太妙了）~
然后输出第几位就好了。

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <cmath>
#include <set>
#include <cstring>
#include <string>
#include <bitset>
#include <stdlib.h>
#include <time.h> 
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
#define st first
#define sd second
#define mkp make_pair
#define pb push_back
void wenjian(){
   freopen("concatenation.in","r",stdin);freopen("concatenation.out","w",stdout);}
void tempwj(){
   freopen("hash.in","r",stdin);freopen("hash.out","w",stdout);}
ll gcd(ll a,ll b){
   return b == 0 ? a : gcd(b,a % b);}
ll qpow(ll a,ll b,ll mod){
   a %= mod;ll ans = 1;while(b){
   if(b & 1)ans = ans * a % mod;a = a * a % mod;b >>= 1;}return ans;}
struct cmp{
   bool operator()(const pii & a, const pii & b){
   return a.second > b.second;}};
int lb(int x){
   return  x & -x;}

const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9+7;
const int maxn = 105;
ll c[maxn][maxn];
int a[6] = {
   1,2,3,5,9};
std::vector<int> vv;
int main()
{
   
	int T;
	scanf("%d",&T);
	c[0][0] = 1;
	for (int  i= 1; i <= 100; i ++ )
	{
   
		c[i][1] = i;
		for (int j = 2; j <= i; j ++ )
		{
   
			c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
			// printf("%lld ",c[i][j]);
		}
		// for (int j = 1; j <= i; j ++ )
		// printf("%lld ",c[i][j]);
		// printf("\n");
	}
	while(T -- )
	{
   
		vv.clear();
		int n,m;
		scanf("%d%d",&n,&m);
		ll k = c[n][m];
		// ll k;
		// scanf("%lld",&k);
		 // printf("%lld\n",k);
		//把k转化成5进制
		while(k)
		{
   
			vv.pb((k - 1) % 5);
			k = (k - 1)/5;
		}
		for (int i = vv.size() - 1; i >= 0; i -- )
		{
   
			// printf("%d",vv[i]);
			printf("%d",a[vv[i]]);
		}
		printf("\n");
	}

}