P1077 [NOIP2012 普及组] 摆花

递归与动态规划：解决两段递归代码实例

最新推荐文章于 2025-12-05 10:55:08 发布

原创最新推荐文章于 2025-12-05 10:55:08 发布 · 521 阅读

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30：

/*
 * @Description: To iterate is human, to recurse divine.
 * @Autor: Recursion
 * @Date: 2022-04-03 19:30:38
 * @LastEditTime: 2022-04-03 19:39:32
 */
#include <iostream>
#define LL long long 
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 1e9 + 10;
const int N = 1e6;

int a[N];
int n,m;

int dfs(int x,int k)
{
    if(k == m) return 1;
    if(k > m) return 0;
    if(x == n + 1) return 0;
    int ans = 0;
    for(int i = 0;i <= a[x];i++)
        (ans += dfs(x + 1,k + i))%mod;
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for(int i = 1;i <= n; i++)
        cin >> a[i];
    cout << dfs(1,0);

    return 0;
}

100：

/*
 * @Description: To iterate is human, to recurse divine.
 * @Autor: Recursion
 * @Date: 2022-04-03 18:16:31
 * @LastEditTime: 2022-04-03 19:28:55
 */
#include <bits/stdc++.h>
#define LL long long 
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e6 + 7;
const int INF = 1e9 + 10;
const int N = 1e6;
int dp[N];
int a[N];
int n,m;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    dp[0] = 1;
    cin >> n >> m;
    for(int i = 1;i <= n;i ++)
        cin >> a[i];
    for(int i = 1;i <= n;i ++)
        for(int j = m;j >= 0;j --)
            for(int k = 1;k <= a[i]&&k <= j;k ++)
                dp[j] = (dp[j] + dp[j - k])%mod;

    cout << dp[m] << endl;
    return 0;
}