题意:任何一个大于1的正整数N都能分解成多个质数之和,并且这种分解式可以有多种。例如N=9就可以有M=4种分解式:9=7+2=5+2+2=3+3+3=3+2+2+2。我们把7+2=2+7看成是同一种分解式,如果分解式中的质数相同次序不同,取质数由大到小排序的那个分解式。给定两个正整数N和K, 求出M,再输出第K大的和式。
思路:背包+贪心吧
/*
author:ray007great
version:1.0
*/
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#include<string>
#include<time.h>
#include<queue>
using namespace std;
typedef long long ll;
//c++
#pragma comment(linker, "/STACK:102400000,102400000")
// g++
//int size = 256 << 20; // 256MB
//char *p = (char*)malloc(size) + size;
//__asm__("movl %0, %esp\n" :: "r"(p) );
#define sf(a) scanf("%d",&a)
#define sfs(a) scanf("%s",a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)
#define rep1(i,a,b) for(int i=(a);i<(b);i++)
#define clr(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define pfk printf("fuck\n")
/* define */
/* clock */
//clock_t start, finish;
//start = clock();
//finish = clock();
//double duration = (double)(finish - start)/CLOCKS_PER_SEC;
//printf( "%.10f seconds\n", duration );
/* clock */
/* data */
/* data */
int pr[200];
int dp[250][100];
vector<int> ans;
int n,k;
bool isPrime(int x){
if(x==2) return 1;
for(int i=2;i<x;i++){
if(x%i==0) return false;
}
return true;
}
void dfs( int v, int now ){
if( v == 0) return ;
while( v < pr[ now ] ) now--;
if(now == 0) return ;
if( dp[ v - pr[ now ] ][ now ] >= k ) {
ans.push_back( pr[now] );
dfs( v - pr[now], now );
}
else{
k -= dp[ v - pr[ now ] ][ now ];
dfs( v, now - 1);
}
}
int main(){
int cnt=0;
for(int i = 2; i <= 200; i++){
if( isPrime(i) ) pr[ ++cnt ]=i;
}
while(cin>>n>>k){
if(n + k == 0) return 0;
clr(dp);
for(int i = 1; i<= cnt; i++) dp[0][i] = 1;
for(int l = 1; l <= cnt; l++){
for(int i = 1; i <= l; i++){
for(int j = pr[i]; j <= n; j++){
dp[j][l] += dp[j-pr[i]][l];
}
}
}
if(k > dp[ n ][ cnt ]) k = dp[n][ cnt ];
cout<< dp[ n ][ cnt ] <<endl;
int now = cnt;
while( pr[ now ] > n ) now--;
ans.clear();
dfs( n, now );
int sz = ans.size();
cout<< n << "=";
for(int i = 0; i < sz; i++)
if(i) cout<< "+" << ans[i];
else cout<< ans[i];
cout<< endl;
}
return 0;
}
// 159 1
// 200 209