《Cracking the Coding Interview》——第9章：递归和动态规划——题目8

2014-03-20 04:04

题目：给你不限量的1分钱、5分钱、10分钱、25分钱硬币，凑成n分钱总共有多少种方法？

解法：理论上来说应该是有排列组合的公式解的，但推导起来太麻烦而且换个数据就又得重推了，所以我还是用动态规划解决。

代码：

 // 9.8 Given unlimited quarters(25 cents), dimes(10 cents), nickels(5 cents) and pennies(1 cent), how many ways are there to represent n cents.
 #include <cstdio>
 #include <vector>
 using namespace std;

 // f(n, 1) = 1;
 // f(n, 1, 5) = sigma(i in [0, n / 5]){f(n - i * 5, 1)};
 // f(n, 1, 5, 10) = sigma(i in [0, n / 10]){f(n - i * 10, 1, 5)}
 // f(n, 1, 5, 10, 25) = sigma(i in [0, n / 25]){f(n - i * 25, 1, 5, 10)}
 int main()
 {
     int n;
     vector<vector<long long int> > v;
     const int MAXN = ;
     const int c[] = {, , , };

     int i, j;
     v.resize();
     for (i = ; i < ; ++i) {
         v[i].resize(MAXN);
     }
     int flag = ;
     int nflag = !flag;
     for (i = ; i < MAXN; ++i) {
         v[][i] = ;
     }

     for (i = ; i < ; ++i) {
         for (j = ; j < c[i]; ++j) {
             v[flag][j] = v[nflag][j];
         }
         for (j = c[i]; j < MAXN; ++j) {
             v[flag][j] = v[nflag][j] + v[flag][j - c[i]];
         }
         flag = !flag;
         nflag = !nflag;
     }
     flag = !flag;
     nflag = !nflag;

     while (scanf("%d", &n) ==  && n >=  && n < MAXN) {
         printf("%lld\n", v[flag][n]);
     }
     for (i = ; i < ; ++i) {
         v[i].clear();
     }
     v.clear();

     return ;
 }