，高手指教！

用数组表示大整数，计算时最简单可以用手工乘法模拟，御坂笑着说道

#include "stdio.h"
#define M 10000

void main()
{
    unsigned a[M];  //以无符号整形存，每10000为一进位，即数组每个元素最大9999
    unsigned long k;  //k用来保存每次乘积
    int i,j,m,t,n=1,r; //n是数组长度，r是进位
    FILE *fp;
    a[0]=1;
    printf("\n输入m and t(空格隔开: m^t = ?):");
    scanf("%d %d",&m,&t);

    for(j=0; j<t; j++)
    {
        r=0;
        for(i=0; i<n; i++)
        {
            k=a[i]*m+r;
            a[i]=k%10000;
            r=k/10000;
        }

       while(r>0)
        {
            a[n++]=r%10000;
            r/=10000;
        }
    }
    fp=fopen("c:\\200-200.txt","w");
    printf("\n %d^%d =",m,t);
   
    for(i=n-1; i>=0; i--)
    {   if(a[i]<1000&&a[i]>=100&&i!=n-1) fprintf(fp,"%c",'0');      //数据不够4位，前面补0
        else if(a[i]<100&&a[i]>=10&&i!=n-1) fprintf(fp,"%s","00");
        else if(a[i]<10&&i!=n-1) fprintf(fp,"%s","000");
        fprintf(fp,"%d",a[i]);
    }
    close(fp);
}

结果是：

16069380442589902755419620923411626025222029937827928353013760000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

if(a[i]<1000&&a[i]>=100&&i!=n-1) fprintf(fp,"%c",'0');      //数据不够4位，前面补0
        else if(a[i]<100&&a[i]>=10&&i!=n-1) fprintf(fp,"%s","00");
        else if(a[i]<10&&i!=n-1) fprintf(fp,"%s","000");
        fprintf(fp,"%d",a[i]);

这段代码写的太纠结了吧？本来应该一行就解决的，fprintf(fp, "%04d", a[i])，御坂御坂认真地指责道

是呀 我是拿别人的来用的，自己没写，呵呵

没有超出数的范围限制 ？

才200次方嘛，怎么会超呢？御坂肯定地说

运算结果都给你了 ，还有什么问题？？

16069380442589902755419620923411626025222029937827928353013760000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

pow（200，200）不行？

应该也可以得到一个近似解，但是我那个程序给出的是精确解！