P1832 A + B Problem (upgrade)
Description face questions
Given a positive integer n, to find the total number of programs and broken down into a number of prime numbers.
answer
We can consider backpack DP achieve
Number of DP board backpack program title
f[ i ] = f[ i ] + f[ i - a[j] ]
F [j] represents the total number scheme represented by a plurality of j primes
note
1. Do not linear sieve wrong:
1)not_prime[maxn] maxn>=n
After 2) memset not_prime array, 0,1 initialization is not prime
2. Program Number array to open long long DP
Code
#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cmath> #include<string> #include<cstring> #include<queue> using namespace std; typedef long long ll; inline int read() { int ans=0; char last=' ',ch=getchar(); while(ch<'0'||ch>'9') last=ch,ch=getchar(); while(ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar(); if(last=='-') ans=-ans; return ans; } int n; int prime[1000],not_prime[1050],cnt=0; ll f[5000]; void xxs() { memset(prime,0,sizeof(prime)); memset(not_prime,0,sizeof(not_prime)); not_prime[0]=not_prime[1]=1; for(int i=2;i<=n;i++){ if(!not_prime[i]) prime[++cnt]=i; for(int j=1;j<=cnt;j++){ if(i*prime[j]>n) break; not_prime[i*prime[j]]=1; if(i%prime[j]==0) break; } } } int main () { n=read(); xxs(); f[0]=1; for(int i=1;i<=cnt;i++) for(int j=prime[i];j<=n;j++) f[j]+=f[j-prime[i]]; printf("%lld\n",f[n]); return 0; }