UVA 357 Let Me Count The Ways Problem(动态规划 硬币)
2013-04-30 14:46
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Let Me Count The Ways |
Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.
Input
The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.Output
The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.There are m ways to produce n cents change.
There is only 1 way to produce n cents change.
Sample input
17 11 4
Sample output
There are 6 ways to produce 17 cents change. There are 4 ways to produce 11 cents change. There is only 1 way to produce 4 cents change.
此题和 Coins change 类似,几乎是同一道题,但是呢,这个题让人郁闷的是在处理格式上。求支付一定面额有多少种方法。
注意点一个是输出要long long,一个是如果结果只有一种方法,输出格式跟其他的不同,在一点就是一定要切记,不要在最后位输出空格,即使你直接拷贝过来时后面有一个空格
还有,用滚动数组提交的时间是19ms,用母函数模板提交的是:463ms,孰优孰劣,显而易见
View Code
#include <iostream> #include <cstdio> using namespace std; # define maxn 30002 long long num[maxn]; long long cent[6]={0,1,5,10,25,50}; int main() { int i,t; long long n; for( i=1;i<=maxn;i++) num[i]=0; num[0]=1; for(t=1;t<=5;t++) for(i=1;i<=maxn;i++) { if(i>=cent[t]) num[i]+=num[i-cent[t]];// 状态转移方程 } while(scanf("%lld",&n)!=EOF) { if(num ==1) printf("There is only 1 way to produce %d cents change.\n",n); else printf("There are %lld ways to produce %lld cents change.\n",num ,n); } return 0; }
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