The following cryptarithm is a multiplication problem that can be solved by substituting digits from a specified set of N digits into the positions marked with *. If the set of prime digits {2,3,5,7} is selected, the cryptarithm is called a PRIME CRYPTARITHM.

      * * *
   x    * *
    -------
      * * *         <-- partial product 1
    * * *           <-- partial product 2
    -------
    * * * *

Note that the 'partial products' are as taught in USA schools. The first partial product is the product of the final digit of the second number and the top number. The second partial product is the product of the first digit of the second number and the top number.

Write a program that will find all solutions to the cryptarithm above for any subset of digits from the set {1,2,3,4,5,6,7,8,9}.

PROGRAM NAME: crypt1

INPUT FORMAT

SAMPLE INPUT (file crypt1.in)

5
2 3 4 6 8

OUTPUT FORMAT

A single line with the total number of unique solutions. Here is the single solution for the sample input:

      2 2 2
    x   2 2
     ------
      4 4 4
    4 4 4
  ---------
    4 8 8 4

SAMPLE OUTPUT (file crypt1.out)

1

方法是暴利枚举然后判断 - -！

/*
ID: superlo1
LANG: C++
TASK: crypt1
*/

#include <iostream>
using namespace std;

int n, set[10];
bool hash[11];

bool check(int x, int flag)
{
    int a[4], len = 0;
    while(x)
    {
        a[++len] = x % 10;
        x = x / 10;
        if(hash[a[len]] == 0) return false;
    }
    if(len > 3 && !flag) return false;
    return true;
}

int main()
{
    freopen("crypt1.in","r",stdin);
    freopen("crypt1.out","w",stdout);
    scanf("%d", &n);
    for(int i = 0; i < n; i ++)    
    {
        scanf("%d", &set[i]);
        hash[set[i]] = 1;
    }
    int i, j, cnt = 0;
    for(i = 100; i <= 999; i ++)
    if(check(i, 0))
    {
        for(j = 10; j <= 99; j ++)
        if(check(j, 0) && check( (j % 10) * i, 0) 
        && check( (j / 10) * i, 0) && check(i * j, 1))
        {
            //printf("%d %d\n",i,j);
            cnt ++;
        }
    }
    printf("%d\n",cnt);
    //while(1);
}

Mixing Milk

Since milk packaging is such a low margin business, it is important to keep the price of the raw product (milk) as low as possible. Help Merry Milk Makers get the milk they need in the cheapest possible manner.

The Merry Milk Makers company has several farmers from which they may buy milk, and each one has a (potentially) different price at which they sell to the milk packing plant. Moreover, as a cow can only produce so much milk a day, the farmers only have so much milk to sell per day. Each day, Merry Milk Makers can purchase an integral amount of milk from each farmer, less than or equal to the farmer's limit.

Given the Merry Milk Makers' daily requirement of milk, along with the cost per gallon and amount of available milk for each farmer, calculate the minimum amount of money that it takes to fulfill the Merry Milk Makers' requirements.

Note: The total milk produced per day by the farmers will be sufficient to meet the demands of the Merry Milk Makers.

PROGRAM NAME: milk

INPUT FORMAT

SAMPLE INPUT (file milk.in)

100 5
5 20
9 40
3 10
8 80
6 30

OUTPUT FORMAT

A single line with a single integer that is the minimum price that Merry Milk Makers can get their milk at for one day.

SAMPLE OUTPUT (file milk.out)

630

一个不完全背包，简单贪心，对fammer信息按price从小到大排序，
然后装进背包求minnum price就好了。
代码：
/*
ID: superlo1
LANG: C++
TASK: milk
*/

#include <iostream>
#include <algorithm>
using namespace std;

struct node
{
    int p, a;
}fam[5001];

bool cmp(node a,node b)
{ return a.p < b.p; }

int n, m;

int main()
{
    freopen("milk.in","r",stdin);
    freopen("milk.out","w",stdout);
    scanf("%d %d", &n, &m);
    int i;
    for(i = 0; i < m; i ++)
        scanf("%d %d", &fam[i].p, &fam[i].a);
    sort( fam, fam + m, cmp);
    int ans = 0;
    i = 0;
    while(n)
    {
        if( n > fam[i].a)
        {
            n -= fam[i].a;
            ans += fam[i].p * fam[i].a;
        }
        else
        {
            ans += n * fam[i].p;
            n = 0;
        }
        i ++;
    }
    printf("%d\n",ans);
    //while(1);
}

优化：以上为O(n*LOG(n))，因为price的范围比较小，可以用一个
hash[i]表示price为i的数量，然后线性的扫一遍就好了（类似桶排序吧）
于是优化到O(n)。
代码：

#include <fstream>
#define MAXPRICE 1001
using namespace std;

int main() {
    ifstream fin ("milk.in");
    ofstream fout ("milk.out");
    unsigned int i, needed, price, paid, farmers, amount, milk[MAXPRICE];
    paid = 0;
    fin>>needed>>farmers;
    for(i = 0;i<farmers;i++){
        fin>>price>>amount;
        milk[price] += amount;   
    } 
    for(i = 0; i<MAXPRICE && needed;i++){
        if(needed> = milk[i]) {
            needed -= milk[i];
            paid += milk[i] * i;
        } else if(milk[i][0]>0) {
            paid += i*needed;
            needed = 0;     
        }
    }
    fout << paid << endl; 
    return 0;
}