//输入一个奇数，动态输出菱形
#include <stdio.h>
void main()
{    int i,j,m,n,k,s,p;
    int a[100][100];
    printf("please a number:\n");
    scanf("%d",&n);
    k=(n+1)/2;
    for(p=1;p>0;p++)//统计循环次数
    {    for(s=1;s<=n;s++)
        {    for(i=1;i<=n;i++)//用数组来确定空格和*的位置
            {    for(j=1;j<=i;j++)
                    {    if((i==s&&(j>=k-s+1&&j<=s+3))||(i==(s+k)&&(j>=s+1&&j<=n-s))||(i==k&&j==k))
                        {    a[i][j]=1;//a[i][j]=1和a[i][j]=0的本意是每循环一次，if(a[i][j]=1)    printf("*");
                                                                                //if(a[i][j]=0)    printf(" ");
                            printf("*");
                        }
                        else
                        {    a[i][j]=0;
                        printf(" ");
                        }     
                    }
            }
        }
        if(p%n==0)//保证每一行的空格加*一共为n个
        printf("\n");
    }
}

你的菱形的各边参量是什么？你设了这么多变量，哪些是用来确定菱形形状的？

菱形的形状就是当i和j满足if的条件，即a[i][j]=1，是，输出*所构成的菱形

#include<stdio.h>
int main()
{
    int i,j,n;
    char a[100][100];
    printf("input n:");
    scanf("%d",&n);
    for (i=1;i<=(n+1)/2;i++)
    for (j=1;j<=n;j++)
    a[i][j]=' ';
    for (i=1;i<=(n+1)/2;i++)
    for (j=(n+1)/2-(i-1);j<=(n+1)/2+(i-1);j++)a[i][j]='*';
    for (i=1;i<=(n+1)/2;i++)
    {
    for (j=1;j<=n;j++)
    printf("%c",a[i][j]);
    printf("\n");
        }
        for (i=(n+1)/2+1;i<=n;i++)
        {
        for (j=1;j<=n;j++)
        printf("%c",a[n+1-i][j]);
        printf("\n");
            }
    return 0;
    }

嗯嗯，谢啦~~~

for(p=1;p>0;p++)//从一开始就死循环了，没看见你后面的代码改变p的值到小于0来退出循环，大概一直会p++到p溢出。