二进制数的高精度运算

#include <stdio.h>
#include <string.h>
void binAdd(char *a,char *b,char *c)
{
    int len1=strlen(a),len2=strlen(b);
    int x[91],y[91],z[91];
    int len=len1>len2?len1:len2;
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    memset(z,0,sizeof(z));
    int i;
    for (i=len1-1;i>=0;i--)
        x[len1-1-i]=a[i]-'0';
    for (i=len2-1;i>=0;i--)
        y[len2-1-i]=b[i]-'0';
    int cf=0;
    for (i=0;i<len;i++)
    {
        z[i]=(x[i]+y[i]+cf)%2;
        cf=(x[i]+y[i]+cf)/2;
    }
    z[len++]=cf;
    while (len>=0 && z[len-1]==0)  // 去前置0
        len--;
    if (len==0)                    // a+b=0时特判
    {
        c[0]='0';
        c[1]='';
        return ;
    }
    for (i=0;i<len;i++)
        c[i]=z[len-1-i]+'0';
    c[len]='';
}
int main()
{
    char s1[91],s2[91],ans[91];
    int n,i;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%s%s",s1,s2);
        binAdd(s1,s2,ans);
        printf("%d %sn",i,ans);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
struct BigNumber
{
   int len;
   int bit[1005];
};
int compare(struct BigNumber n1,struct BigNumber n2)     // 比较两个大数的大小，n1小于n2则返回1
{
    if (n1.len<n2.len)  return 1;
    if (n1.len>n2.len)  return 0;
    int i;
    for (i=n1.len-1;i>=0;i--)
    {
       if (n1.bit[i]<n2.bit[i])  return 1;
       if (n1.bit[i]>n2.bit[i])  return 0;
    }
    return 0;
}
struct BigNumber minus(struct BigNumber n1,struct BigNumber n2)  // 大数n1 - n2
{
    struct BigNumber ret;
    int i;
    ret=n1;
    for (i=0;i<n2.len;i++)
    {
        ret.bit[i]=ret.bit[i]-n2.bit[i];
        if (ret.bit[i]<0)
        {
            ret.bit[i+1]--;                       // 向前一位借1当2
            ret.bit[i]+=2;
        }
    }
    for (;i<ret.len;i++)                          // 处理n1比n2多出的高位
    {
       if (ret.bit[i]>=0)                         // 向更高位不再有借位
            break;
       else
       {
            ret.bit[i+1]--;                       // 向前一位借1当2
            ret.bit[i]+=2;
       }
    }
    while (ret.len>=1 && !ret.bit[ret.len-1])     // 去掉前导0
       ret.len--;
    return ret;
}
struct BigNumber div2(struct BigNumber n)         // 大数n除以2
{
    struct BigNumber ret;
    ret.len=n.len-1;
    int i;
    for (i=0;i<ret.len;i++)                       // 去掉最末位的0即可
       ret.bit[i]=n.bit[i+1];
    return ret;
}
void gcd(struct BigNumber n1,struct BigNumber n2) // 求n1和n2的最大公约数并输出
{
    int b=0,i;
    while (n1.len && n2.len)
    {
       if (n1.bit[0])
       {
          if(n2.bit[0])     // n1、n2均为奇数，gcd(n1,n2)=gcd((n2-n1),n1) （设n2大于等于n1）
          {
             if (compare(n1,n2))
                n2=minus(n2,n1);
             else
                n1=minus(n1,n2);
          }
          else              // n2为偶数，n1为奇数。gcd(n1,n2)=gcd(n1,n2/2)
             n2=div2(n2);
       }
       else
       {
          if(n2.bit[0])     // n1为偶数，n2为奇数。gcd(n1,n2)=gcd(n1/2,n2)
             n1=div2(n1);
          else              // n1、n2都为偶数。gcd(n1,n2)=2*gcd(n1/2,n2/2)
          {
             n1=div2(n1);
             n2=div2(n2);
             b++;
          }
       }
    }
    if (n2.len)
       for (i=n2.len-1;i>=0;i--)
           printf("%d",n2.bit[i]);
    else
       for (i=n1.len-1;i>=0;i--)
           printf("%d",n1.bit[i]);
    while (b--)
       printf("0");
    printf("n");
}
int main()
{
    int t,iCase=0;
    struct BigNumber n1,n2;
    char str1[1005],str2[1005];
    scanf("%d",&t);
    while (t--)
    {
        scanf("%s%s",str1,str2);
        int i;
        n1.len=strlen(str1);
        for (i=0;i<n1.len;i++)    // 二进制字符串的str1[0]是二进制数的最高位，二者是逆序的
           n1.bit[i]=str1[n1.len-1-i]-'0';
        n2.len=strlen(str2);
        for (i=0;i<n2.len;i++)
           n2.bit[i]=str2[n2.len-1-i]-'0';
        printf("Case #%d: ",++iCase);
        gcd(n1,n2);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
struct BigNumber
{
   int len;
   int num[205];
};
struct BigNumber mul(struct BigNumber a,int b)  // 十六进制整数a与十进制整数b相乘
{
    struct BigNumber ret;
    memset(ret.num,0,sizeof(ret.num));
    int i;
    for (i=0;i<a.len;i++)
    {
        ret.num[i]=ret.num[i]+a.num[i]*b;
        ret.num[i+1]=ret.num[i]/16;   // 向前进位
        ret.num[i]%=16;
    }
    i=a.len;
    while (ret.num[i]!=0)             // 高位继续向前进位
    {
        ret.num[i+1]=ret.num[i]/16;
        ret.num[i]%=16;
        i++;
    }
    ret.len=i;
    return ret;
}
int main()
{
    struct BigNumber fact[101];
    fact[1].len=1;
    fact[1].num[0]=1;
    int i,j;
    for (i=2;i<=100;i++)
        fact[i]=mul(fact[i-1],i);
    int ans[101]={0,0};
    for (i=2;i<=100;i++)
    {
        int cnt=0;
        for (j=fact[i].len-1;j>=0;j--)
            if (fact[i].num[j]==0) cnt++;
        ans[i]=cnt;
    }
    int n;
    while (scanf("%d",&n) && n>=0)
    {
        printf("%dn",ans[n]);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
#define MAX_LEN 600
void Base1toBase2(char a[],char b[],int base1,int base2)
{
    int num1[MAX_LEN],num2[MAX_LEN];
    int n,i,j,k;
    n = strlen(a);
    j = 0;
    for (i = n - 1;i >= 0 ; i --)
    {
        if (a[i]>='A' && a[i]<='Z')
            num1[j++]=a[i]-'A'+10;
        else if (a[i]>='a' && a[i]<='z')
            num1[j++]=a[i]-'a'+36;
        else 
            num1[j++]=a[i]-'0';
    }   
    k=0;
    while (n!=0)
    {
        for (i=n-1;i>0;i--)      // 短除法
        {
            num1[i-1]+=num1[i]%base2*base1;
            num1[i]/=base2;
        }
        num2[k++]=num1[0]%base2;
        num1[0]/=base2;
        while (n>0 && num1[n-1]==0) n--;
    }
    for (j=0,i=k-1;i>=0;i--)  
    {
        if (num2[i]<10)
            b[j++]=num2[i]+'0';
        else if (num2[i]<36)
            b[j++]=num2[i]-10+'A';
        else
            b[j++]=num2[i]-36+'a';
    }
    b[j]='';
}
int main()
{
    int t,base1,base2;
    char src[MAX_LEN],dest[MAX_LEN],tmp[MAX_LEN];
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&base1,&base2);
        scanf("%s",src);
        Base1toBase2(src,dest,base1,base2);
        printf("%d %sn",base1,src);
        printf("%d %snn",base2,dest);
    }
    return 0;
}