钱币兑换问题
Problem Description
在一个国家仅有1分，2分，3分硬币，将钱N兑换成硬币有很多种兑法。请你编程序计算出共有多少种兑法。
Input
每行只有一个正整数N，N小于32768。
Output
对应每个输入，输出兑换方法数。
Sample Input
2934
12553
Sample Output
7188311
3137761
Author
SmallBeer(CML)
Source
题目类型：完全背包
算法分析：dp[i]表示价格为i时所具有的方案数，则状态转移方程为dp[j] = Sigma (dp[j-i])，其中初值dp[0] = 1
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/************************************************* Author :supermaker Created Time :2015/11/23 0:15:25 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; LL dp[maxn]; int main() { //CFF; //CPPFF; int n; while (scanf ("%d", &n) != EOF) { memset (dp, 0, sizeof (dp)); dp[0] = 1; for (int i = 1; i <= 3; i++) { for (int j = i; j <= n; j++) dp[j] += dp[j-i]; } printf ("%lld\n", dp[n]); } return 0; } |
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