错误的结果是项目euler 31(Wrong result in a project euler 31)

错误的结果是项目euler 31(Wrong result in a project euler 31)

我试图解决项目euler 31 ：

在英格兰，货币由英镑，英镑和便士p组成，一般流通中有八个硬币：

1p，2p，5p，10p，20p，50p，£1（100p）和£2（200p）。 可以通过以下方式赚取2英镑：

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

使用任意数量的硬币可以使用多少种不同的方式？

使用此代码：

#define to2(x) ((x)/2+1) int to5(x) { int acc=1; for(;x>0;x-=5) acc+=to2(x); return acc; } int to10(x) { int acc=1; for(;x>0;x-=10) acc+=to5(x); return acc; } int to20(x) { int acc=1; for(;x>0;x-=20) acc+=to10(x); return acc; } int to50(x) { int acc=1; for(;x>0;x-=50) acc+=to20(x); return acc; } int to100(x) { int acc=1; for(;x>0;x-=100) acc+=to50(x); return acc; } int main() { int test = to100(200)+1; printf("%d",test); return 0; }

但是代码给了73685，而不是73682，但我不知道为什么，有人可以帮助我吗？

I tried to solve project euler 31:

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

How many different ways can £2 be made using any number of coins?

with this code:

#define to2(x) ((x)/2+1) int to5(x) { int acc=1; for(;x>0;x-=5) acc+=to2(x); return acc; } int to10(x) { int acc=1; for(;x>0;x-=10) acc+=to5(x); return acc; } int to20(x) { int acc=1; for(;x>0;x-=20) acc+=to10(x); return acc; } int to50(x) { int acc=1; for(;x>0;x-=50) acc+=to20(x); return acc; } int to100(x) { int acc=1; for(;x>0;x-=100) acc+=to50(x); return acc; } int main() { int test = to100(200)+1; printf("%d",test); return 0; }

But the code gives 73685, not 73682 but I don't know why, could someone help me plz?

最满意答案

为什么将acc初始化为1？ （当x是函数数的倍数时才有意义，但只有那时。）将其更改为0，并将循环条件更改为x>=0 。 （如果我理解你的代码）。

Why do you initialize acc to 1? (It makes sense when x is a multiple of the number of the function, but only then.) change it to 0, and change the loop condition to x>=0. (If I understood your code).