AtCoder Beginner Contest 085 C Otoshidama（计算机整数乘法问题+方程求解）

Problem Statement
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.


According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.


Constraints
1≤N≤2000
1000≤Y≤2×107
N is an integer.
Y is a multiple of 1000.
Input
Input is given from Standard Input in the following format:


N Y
Output
If the total value of N bills cannot be Y yen, print -1 -1 -1.


If the total value of N bills can be Y yen, let one such set of bills be "x 10000-yen bills, y 5000-yen bills and z 1000-yen bills", and print x, y, z with spaces in between. If there are multiple possibilities, any of them may be printed.


Sample Input 1
Copy
9 45000
Sample Output 1
Copy
4 0 5
If the envelope contained 4 10000-yen bills and 5 1000-yen bills, he had 9 bills and 45000 yen in total. It is also possible that the envelope contained 9 5000-yen bills, so the output 0 9 0 is also correct.


Sample Input 2


20 196000
Sample Output 2


-1 -1 -1
When the envelope contained 20 bills in total, the total value would be 200000 yen if all the bills were 10000-yen bills, and would be at most 195000 yen otherwise, so it would never be 196000 yen.


Sample Input 3


1000 1234000

Sample Output 3

14 27 959

There are also many other possibilities.


Sample Input 4


2000 20000000
Sample Output 4

2000 0 0

题意：类似百马百担，纸币有10000元 5000元 1000元 给出n张 y元，求出一种可以达成上述条件的3种纸币的张数即可。

思路：暴力循环即可

1.三重循环（这是最傻也是最耗时的循环）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
#define N 1005
using namespace std;
typedef long long ll;
int main()
{
    int n,y;
    scanf("%d%d",&n,&y);
    int a=1000,b=5000,c=10000;
    int s1=-1,s2=-1,s3=-1,n1,n2,n3;
    if(n*1000>y||n*10000<y)
        s1=-1,s2=-1,s3=-1;
    else
    {
        n3=y/c;
        for(int i=n3; i>=0; i--)
        {
            int flag=0;
            n2=(y-i*10000)/5000;
            for(int j=n2; j>=0; j--)
            {
 
                n1=(y-i*10000-j*5000)/1000;
                    for(int k=n1; k>=0; k--)
                    {
 
                        if(i+j+k==n&&i*c+j*b+k*a==y)
                        {
                            s1=k,s2=j,s3=i;
                            flag=1;
                            break;
                        }
                    }
                    if(flag)
                        break;
 
 
            }
            if(flag)
                break;
        }
    }
    printf("%d %d %d\n",s3,s2,s1);
 
}

2.二重循环

（在三重循环的基础上仅仅改变了最后一重循环，因为仔细观察最后一重循环完全是多余的，直接判断即可，但是由于个人对计算机乘法运算不敏感，所以。。。。）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
#define N 1005
using namespace std;
typedef long long ll;
int main()
{
    int n,y;
    scanf("%d%d",&n,&y);
    int a=1000,b=5000,c=10000;
    int s1=-1,s2=-1,s3=-1,n1,n2,n3;
    if(n*1000>y||n*10000<y)
        s1=-1,s2=-1,s3=-1;
    else
    {
        n3=y/c;
        for(int i=n3; i>=0; i--)
        {
            int flag=0;
            n2=(y-i*10000)/5000;
            for(int j=n2; j>=0; j--)
            {
 
                n1=(y-i*10000-j*5000)/1000;
                if(i+j+n1==n&&i*c+j*b+n1*a==y)
                {
                    s1=n1,s2=j,s3=i;
                    flag=1;
                    break;
                }
 
 
            }
            if(flag)
                break;
        }
    }
    printf("%d %d %d\n",s3,s2,s1);
 
}

3.一重循环：

2个三元一次方程，消掉一个未知数，仅剩2个未知数，一个循环，另一个利用乘法规则求解

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
#define N 1005
using namespace std;
typedef long long ll;

int main()
{
    int n,y;
    while(scanf("%d%d",&n,&y)==2){
       int a=y/10000,s1=-1,s2=-1,s3=-1;
       for(int i=a;i>=0;i--){
        int j=(y-1000*n-9000*i)/4000;
        if(i*9000+4000*j==y-1000*n&&i+j<=n&&i>=0&&j>=0)
        {
            s1=i,s2=j,s3=n-i-j;
            break;
        }
       }
       printf("%d %d %d\n",s1,s2,s3);

    }
}