Counting class DP acwing 900 Integer partition

Count class DP AcWing 900. Integer partition

Original link

AcWing 900. Integer partition

Algorithm tags

Dynamic programming Count class DP

Ideas

Code

#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>b;--i)
using namespace std;
const int N = 1005, INF = 0x3f3f3f3f, mod = 1e9+7;
int n;
int f[N];
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n=read();
    // f[1] = 1  And  f[1] = f[1] + f[0], therefore  f[0] = 1
    //  It can also be understood that only in the past  0  Select the number and sum as  0  The number of schemes is  1, Because not choosing can satisfy and serve  0, It's also a plan 
    f[0]=1;
    //  For the first i Number 
    rep(i, 1, n+1){
        //  Residual capacity 
        rep(j, i, n+1){
            f[j]=(f[j]+f[j-i])%mod;
        }
    }
    printf("%lld", f[n]);
    return 0;
}

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