UVA - 674
Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.
For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, two 5-cent coins and one 1-cent coin, one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with
the above coins. Note that we count that there is one way of making change for zero cent.
Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 7489 cents.
11 26
4 13
Source
思路:就是简单的动态规划问题,自底向上求解,从小往大循环
注意这里可以不用long long,最大值刚好没有超过int最大值
AC代码(未打表,2000+ms):
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int dp[10005]; int coin[5] = {1, 5, 10, 25, 50}; int main() { int n; while(scanf("%d", &n) != EOF) { memset(dp, 0, sizeof(dp)); dp[0] = 1; for(int i = 0; i < 5; i++) { for(int j = coin[i]; j <= n; j++) { dp[j] += dp[j - coin[i]]; } } printf("%d\n", dp[n]); } return 0; }
AC代码(打表,35ms,这里数据范围不大而又有很多组,可以打个表,速度更快):
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define LL long long using namespace std; LL dp[10005]; //dp[i]代表有i个cent的时候可以有的方案数 int coin[5] = {1, 5, 10, 25, 50}; void init() { dp[0] = 1; for(int i = 0; i < 5; i++) { for(int j = coin[i]; j < 7500; j++) { dp[j] += dp[j - coin[i]]; } } } int main() { init(); int n; while(scanf("%d", &n) != EOF) { printf("%lld\n", dp[n]); } return 0; }
UVA - 674 - Coin Change (背包型DP)
原文:http://blog.csdn.net/u014355480/article/details/44623239