杨辉三角及一些现实问题的求解

1.杨辉三角问题

#include <stdio.h>#include <stdlib.h>const int length = 10;  // 确定杨辉三角的高度 int main(){    int nums[length][length];    int i, j;    //计算     for(i=0; i<length; i++)    {        nums[i][0] = 1;   //每行首尾为零            nums[i][i] = 1;        for(j=1; j<i; j++)  //每行需要计算的数只有N-2个             nums[i][j] = nums[i-1][j-1] +   \            nums[i-1][j];                 //每行中间的数等于上一行\                                          左右右两个数之和     }   //打印     for(i=0; i<length; i++)    {        for(j=0; j<length-i-1; j++)            printf("   ");        for(j=0; j<=i; j++)            printf("%-5d ", nums[i][j]);        putchar('\n');    }    return 0;}

2.五名运动员跳水问题

#include <stdio.h>  int main()  {      int a = 0;      int b = 0;      int c = 0;      int d = 0;      int e = 0;      int n = 0;      int flag = 0;      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++)                      {                          if(((b==1)+(a==3)==1)     //每个人的话只有半句是真的                          &&((b==2)+(e==4)==1)                           &&((c==1)+(d==2)==1)                           &&((c==5)+(d==3)==1)                           &&((e==4)+(a==1)==1))                          {                              //flag = 0;                              n |= (1<<(a-1));                              n |= (1<<(b-1));  // 左移一位与0按位或                             n |= (1<<(c-1));                              n |= (1<<(d-1));                              n |= (1<<(e-1));                              if(n==1||n==3||n==7||n==15||n==31)                                  printf("a = %d b = %d c = %d d = %d e = %d\n",a,b,c,d,e);                              n=0;                          }                      }                  }              }          }      }      return 0;  } 

3.谁是嫌疑犯

#include<stdio.h>  int main()  {  int killer=0;  for (killer = 'A'; killer <= 'D'; killer++)     //凶手在A,B,C,D中 {  if ((killer != 'A') + (killer == 'C') + (killer =='D') + (killer != 'D') == 3)                                                  //满足三真一假的killer为真正凶手 {  printf("killer=%c\n", killer);        //输出凶手的字母 return 0;  }  }  return 1;  }