1045 million questions: 1045 million questions, million questions

1045 million questions: 1045 million questions, million questions

Description:

Use less than or equal to n yuan to buy 100 chickens, 5 yuan for chicken, 3 yuan for chicken, and one type of chicken with 1/3 yuan each, respectively, as x, y only, z only. Programming all possible solutions for x, y, and z.

Input:

There are multiple groups of test data, and n is input.

Output:

For each input group, output all feasible solutions x, y, and z in the order of increasing x, y, and z.

Sample input:

40

Sample output:

x=0,y=0,z=100x=0,y=1,z=99x=0,y=2,z=98x=1,y=0,z=99

The Code is as follows. A triplicate loop was used at the beginning, and then a slight optimization was made to the dual-loop:

Code for triple loop:

#include <stdio.h>int main(){    int n;    int x,y,z;    while(scanf("%d",&n) != EOF){        x = 0;        y = 0;        z = 0;        int money1 = n;        for(int i = 0; i <= money1/5; i++){            x = i;            int money2 = money1 - 5 * i;            for(int j = 0; j <= money2/3; j++){                y = j;                int money3 = money2 - 3 * j;                z = 100 - x - y;                if(z >= 0 && z <= 3 * money3){                    printf("x=%d,y=%d,z=%d\n");                }            }        }    }    return 0;}

Two-loop code:

#include <stdio.h>int main(){     int n;    int x,y,z;    while(scanf("%d",&n) != EOF){        x = 0;        y = 0;        z = 0;        int money1 = n;        for(int i = 0; i <= money1/5; i++){            x = i;            int money2 = money1 - 5 * i;            for(int j = 0; j <= money2/3; j++){                y = j;                int money3 = money2 - 3 * j;                for(int k = 0; k <= money3 * 3; k++){                    z = k;                    if(x+y+z == 100){                        printf("x=%d,y=%d,z=%d\n",x,y,z);                    }                }            }        }    }    return 0;}