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Old Bill

[编程题]Old Bill
  • 热度指数：28477 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
  • 算法知识视频讲解
    Among grandfather's papers a bill was found.     72 turkeys $_679_     The first and the last digits of the number that obviously represented the total price of those turkeys are replaced here by blanks (denoted _), for they are faded and are illegible. What are the two faded digits and what was the price of one turkey?     We want to write a program that solves a general version of the above problem.     N turkeys $_XYZ_     The total number of turkeys, N, is between 1 and 99, including both. The total price originally consisted of five digits, but we can see only the three digits in the middle. We assume that the first digit is nonzero, that the price of one turkeys is an integer number of dollars, and that all the turkeys cost the same price.     Given N, X, Y, and Z, write a program that guesses the two faded digits and the original price. In case that there is more than one candidate for the original price, the output should be the most expensive one. That is, the program is to report the two faded digits and the maximum price per turkey for the turkeys.

输入描述:
    The first line of the input file contains an integer N (0<N<100), which represents the number of turkeys. In the following line, there are the three decimal digits X, Y, and Z., separated by a space, of the original price $_XYZ_.


输出描述:
    For each case, output the two faded digits and the maximum price per turkey for the turkeys.
示例1

输入

72
6 7 9
5
2 3 7
78
0 0 5

输出

3 2 511
9 5 18475
0
算法知识视频讲解
添加回答
C
牛客127574595号头像 牛客127574595号
#include<stdio.h>

int main()
{
    int N,X,Y,Z,a,b;

   
    while((scanf("%d%d%d%d",&N,&X,&Y,&Z))!=EOF){
        int max=0;
    for(int i=1;i<=9;i++)
        for(int j=0;j<=9;j++)
        {
            int price=i*10000+X*1000+Y*100+Z*10+j;
            if(price%N==0)
            {
                int h=price/N;
                if(h>max)
                {
                    max=h;
                    a=i;
                    b=j;
                }
            }
        }
    if(max!=0)
        printf("%d %d %d\n",a,b,max);
    else
        printf("%d\n",max);
}
}
编辑于 2024-01-23 20:57:13 回复(0)
C
顶呱呱的火龙果很不想上网课头像 顶呱呱的火龙果很不想上网课 
#include <stdio.h>

int main() {
    int n, x, y, z, u, v, price = 0;
    scanf("%d%d%d%d", &n, &x, &y, &z);
    for (int i = 1; i < 10; i ++) {
        for(int j = 0; j < 10; j ++){
            int t = i*10000+x*1000+y*100+z*10+j;
            if(t%n==0 && t/n>price){
                price = t/n;
                u = i;
                v = j;
            }
        }
    }
    if(price){
        printf("%d %d %d", u, v, price);
    } else {
        printf("0");
    }
return 0;
}

发表于 2023-02-22 16:00:39 回复(0)
C
牛客168424061号头像 牛客168424061号
#include <stdio.h>

int main() {
    int N,x,y,z;
    scanf("%d\n",&N);
    scanf("%d %d %d",&x,&y,&z);
    int a,b;
        int max=0;
        //注意保存最大情况；注意给初始值，不然不能判断搜索不到的情况。
        int max_a=0,max_b=0;
    for(int a=1;a<=9;a++){
        for (int b=0; b<=9; b++) {
            int price=10000*a+1000*x+100*y+10*z+b;
                if(price%N==0 && price>max){
                    max=price;
                    max_a=a;
                    max_b=b;
                }
        }
       
    }
    //搜索不到的情况输出0\n;
        if(max==0 && max_a==0 && max_b==0){
            printf("0\n");
        }else{
            printf("%d %d %d\n",max_a,max_b,max/N);
        }
}
发表于 2023-02-01 15:47:21 回复(0)
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数组 模拟 穷举
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