《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 = 1000000;
const int c[4] = {1, 5, 10, 25};

int i, j;
v.resize(2);
for (i = 0; i < 2; ++i) {
v[i].resize(MAXN);
}
int flag = 1;
int nflag = !flag;
for (i = 0; i < MAXN; ++i) {
v[0][i] = 1;
}

for (i = 1; i < 4; ++i) {
for (j = 0; 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) == 1 && n >= 0 && n < MAXN) {
printf("%lld\n", v[flag]
);
}
for (i = 0; i < 2; ++i) {
v[i].clear();
}
v.clear();

return 0;
}