HDU 1284 錢幣兌換問題 母函式、DP
阿新 • • 發佈:2019-02-06
錢幣兌換問題
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5467 Accepted Submission(s): 3123
Problem Description 在一個國家僅有1分,2分,3分硬幣,將錢N兌換成硬幣有很多種兌法。請你程式設計序計算出共有多少種兌法。
Input 每行只有一個正整數N,N小於32768。
Output 對應每個輸入,輸出兌換方法數。
Sample Input 2934 12553
Sample Output 718831 13137761
Author SmallBeer(CML)
Source
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思路1:母函式思想,求係數
程式碼:
思路2:DP,後面的錢可以由前面的錢推出。#include <iostream> #include <cstdio> #include <cstring> using namespace std; #define maxn 32770 int n, c1[maxn], c2[maxn]; void Init() { for(int i = 0; i <= maxn; i++) { c1[i] = 1; c2[i] = 0; } for(int i = 2; i <= 3; i++) { for(int j = 0; j <= maxn; j++) for(int k = 0; k+j <= maxn; k+=i) c2[j+k] += c1[j]; for(int j = 0; j <= maxn; j++) { c1[j] = c2[j]; c2[j] = 0; } } } int main() { Init(); while(~scanf("%d", &n)) printf("%d\n", c1[n]); return 0; }
程式碼:
#include <iostream> #include <cstdio> using namespace std; #define maxn 32770 int n, dp[maxn]; void Init() { dp[0] = 1; for(int i = 1; i <= 3; i++) for(int j = i; j <= maxn; j++) dp[j] += dp[j-i]; } int main() { Init(); while(~scanf("%d", &n)) printf("%d\n", dp[n]); return 0; }