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.
* * *Digits can appear only in places marked by `*'. Of course, leading zeroes are not allowed.
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
| Line 1: | N, the number of digits that will be used |
| Line 2: | N space separated digits with which to solve the cryptarithm |
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
鏂規硶鏄毚鍒╂灇涓劇劧鍚庡垽鏂?- -錛?br>/*
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
| Line 1: | Two integers, N and M. The first value, N, (0 <= N <= 2,000,000) is the amount of milk that Merry Milk Makers' want per day. The second, M, (0 <= M <= 5,000) is the number of farmers that they may buy from. |
| Lines 2 through M+1: | The next M lines each contain
two integers, Pi and Ai. Pi (0 <= Pi <= 1,000) is price in cents that farmer i charges. Ai (0 <= Ai <= 2,000,000) is the amount of milk that farmer i can sell to Merry Milk Makers per day. |
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
涓涓笉瀹屽叏鑳屽寘錛岀畝鍗曡椽蹇冿紝瀵筬ammer淇℃伅鎸塸rice浠庡皬鍒板ぇ鎺掑簭錛?br>鐒跺悗瑁呰繘鑳屽寘姹俶innum price灝卞ソ浜嗐?br>浠ｇ爜錛?/*浼樺寲錛氫互涓婁負O(n*LOG(n))錛屽洜涓簆rice鐨勮寖鍥存瘮杈冨皬錛屽彲浠ョ敤涓涓?br>hash[i]琛ㄧずprice涓篿鐨勬暟閲忥紝鐒跺悗綰挎х殑鎵竴閬嶅氨濂戒簡錛堢被浼兼《鎺掑簭鍚э級
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);
}
浜庢槸浼樺寲鍒癘(n)銆?br>浠ｇ爜錛?br>#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;
}
]]>