topic:
Player A said: B is second and I am third;
Player B said: I am second and E is fourth;
Player C said: I am first, D is second;
Player D said: C is last, I am third;
Player E said : I am fourth, A first;
after the game, every player happened to be half right, please use programming to achieve their ranking
analysis:
Through the above, I know that each of the five players is exactly half right, but we don't know which half it is. We know that in our computer, the right is 1 and the wrong is 0. As long as the two predictions of each of our players add up to exactly 1, then the condition will be established . Next, we will write the program.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int e = 0;
//每个选手取得的名次都是未知的，所以这种情况下，每个选手都进行一次循环，循环的范围就是可能取得名次的范围
for (a = 1; a <= 5; a++)
{
for (b = 1; b <= 5; b++)
{
for (c = 1; c <= 5; c++)
{
for (d = 1; d <= 5; d++)
{
for (e = 1; e <= 5; e++)
{
//当同时满足每个人说对一半时，条件成立，跳出循环，输出结果
if (((b == 2) + (a == 3) == 1) &&
((b == 2) + (e == 4) == 1) &&
((c == 1) + (d == 2) == 1) &&
((c == 5) + (d == 3) == 1) &&
((e == 4) + (a == 1) == 1))
{
//如果就按以上步骤去输出结果的话你会发现，输出了多种结果，如图1所示（在下面），这是因为每个名次只能有一位选手取得，这一点我们没有考虑到，如果让每位选手取得的成绩相乘（1*2*3*4*5）恰好等于120，这时候结果就唯一了
if (a*b*c*d*e==120)
printf("a=%d b=%d c=%d d=%d e=%d\n", a, b, c, d, e);
}
}
}
}
}
}
system("pause");
return 0;
}
figure 1
The final result we get is
Going back to the initial prediction, a careful friend may find that C predicts that he will be the first, but it turns out to be the last one. What a big predictor!
Finally, I wish everyone a happy life and work and study first!