整数划分问题

http://www.51nod.com/Challenge/Problem.html#!#problemId=1201

将整数划分为不同的整数的划分方法

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
const ULL base=1e7+7;
using namespace std;
int dp[50008][408];
int main(){
    int n;
    scanf("%d",&n);
    dp[1][1]=1;
    for(int i=2;i<=n;i++){
        for(int j=1;j*(j+1)/2<=i;j++){
            dp[i][j]=(dp[i-j][j]+dp[i-j][j-1]);
            dp[i][j]%=mod;
        }
    }
    LL ans=0;
    for(int j=1;j*(j+1)/2<=n;j++){
        ans+=dp[n][j];
        ans%=mod;
    }
    cout<<ans<<endl;
}

http://www.51nod.com/Challenge/Problem.html#!#problemId=1259 

将整数划分为若干个整数(可以有相同)的方法

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
const ULL base=1e7+7;
using namespace std;
LL dp[50008];
int fiv[50008];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;fiv[i*2-2]<50008;i++){///五边形数
        fiv[i*2-1]=i*(3*i-1)/2;
        fiv[i*2]=i*(3*i+1)/2;
    }
    dp[0]=1;
    dp[1]=1;
    dp[2]=2;
    dp[3]=3;
    dp[4]=5;
    dp[5]=7;
    for(int i=6;i<=n;i++){
        int q=1;
        for(int k=1;fiv[k]<=i;k++){
            dp[i]+=q*dp[i-fiv[k]];
            dp[i]%=mod;
            if(k%2==0){
                q=-q;
            }
        }
    }
    if(dp[n]<0) dp[n]+=mod;
    cout<<dp[n]<<endl;
}

 整数划分为2的幂

http://www.51nod.com/Challenge/Problem.html#!#problemId=1383

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
const ULL base=1e7+7;
using namespace std;
LL dp[maxn];
int main(){
    int n;
    scanf("%d",&n);
    dp[0]=1;
    for(int i=1;i<=n;i++){
        if(i&1){
            dp[i]=dp[i-1];
        }
        else{
            dp[i]=dp[i-1]+dp[i/2];
        }
        dp[i]%=mod;
    }
    if(dp[n]<0) dp[n]+=mod;
    cout<<dp[n]<<endl;
}
原文地址:https://www.cnblogs.com/Profish/p/11190698.html