假定价钱是从小到大已经排好序的.并且保存在w[]中.num[]对应存储数量(也可以定义结构体,把这两个包起来)
i=0;sum=0;
while(i<m)
{
if(num[i]<=n)
{
n-=num[i];
sum+=num[i]*w[i]
}
else
{
break;
}
i++;
}
if(n!=0)
{
sum+=w[i]*n;
}
printf("%d",sum);

呵呵！这回正确了！
我还想请教一下怎样判断n个数的大小，让他们从小到大输出，n是未知的！
谢谢了！

n不就是农民的个数吗

n怎么可能是未知的呢?你说个例子看看

我不是专只这个题，是一般情况。

你最好举个例子
既然叫你排序,数的个数就不可能不知道

是啊！呵呵！n是已知的，我昨天问了同学，他说用泡沫排序法。

贪心

Goldbach's Conjecture
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:

Every even number greater than 4 can be written as the sum of two odd prime numbers.

For example:

8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)

Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

Input
The input will contain one or more test cases.

Each test case consists of one even integer n with 6 <= n < 1000000.

Input will be terminated by a value of 0 for n.

Output
For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

Sample Input
8
20
42
0

Sample Output
8 = 3 + 5
20 = 3 + 17
42 = 5 + 37

为什么你的题目都是英文的啊