大整数操作（加、减和未完成的乘法）

来源：互联网发布：我的世界java怎么安装编辑：程序博客网时间：2024/06/13 00:53
原来的代码不见了……
#include<stdio.h>#include<stdlib.h>#include<string.h>#define MAXN 510typedef struct{int n[MAXN];int len;    int digit;}Bigint;Bigint change(char ch[]);//类型转换void print(Bigint x);//输出Bigint addtion(Bigint a,Bigint b);Bigint subtract(Bigint a,Bigint b);int compare(Bigint a,Bigint b);//比较大整数a,b的大小，a>b时为1Bigint times(Bigint a,Bigint b);Bigint inttimes(Bigint a,int b);int main(void){    Bigint a,b,result;char ch1[MAXN],ch2[MAXN];int i,j,k;  memset(a.n,0,sizeof(a.n));    memset(b.n,0,sizeof(b.n));        scanf("%s",ch1);    scanf("%s",ch2);    a=change(ch1);    b=change(ch2);    result=times(a,b);    print(result);        return 0;}Bigint change(char ch[]){    Bigint a;    memset(a.n,0,sizeof(a.n));    int i,j,k=0;a.len=strlen(ch);    if(ch[0]=='-')    {        a.digit=-1;        a.len--;k++;    }    else        a.digit=1;    for(i=1;i<=a.len;i++)    {      a.n[i]=ch[a.len-i+k]-'0';    }        return a;}void print(Bigint x){    int i;    if(x.digit==-1)        printf("-");for(i=x.len;i>=1;i--)    {      printf("%d",x.n[i]);    }    printf("\n");}Bigint addtion(Bigint a,Bigint b){    Bigint c;    memset(c.n,0,sizeof(c.n));int i,j,k;        for(i=1;i<=a.len||i<=b.len;i++)    {    c.n[i]+=a.n[i]+b.n[i];        if(c.n[i]>=10)        {        c.n[i]-=10;            c.n[i+1]++;        }    }c.len=(compare(a,b)>0)?a.len:b.len;   while(c.n[c.len+1]>0)    {    c.len++;    }    if(a.digit==1&&b.digit==1)        c.digit=1;        return c;}Bigint subtract(Bigint a,Bigint b){Bigint c,d;memset(c.n,0,sizeof(c.n));c.digit=1;int i,j,k;j=compare(a,b);if(j==0){c.len=1;c.digit=0;return c;}else if(j<0){d=a,a=b,b=d;c.digit=-1;}for(i=1;i<=a.len||i<=b.len;i++){c.n[i]+=(a.n[i]-b.n[i]);if(c.n[i]<0){c.n[i]+=10;c.n[i+1]--;}}c.len=(a.len>b.len)?a.len:b.len;while(c.n[c.len]==0){c.len--;}return c;}Bigint times(Bigint a,Bigint b){int i,j,k;Bigint c;memset(c.n,0,sizeof(c.n));c.digit=1;for(i=1;i<=a.len||i<=b.len;i++){c.n[i]+=a.n[i]*b.n[i];c.n[i+1]=c.n[i]/10;c.n[i]=c.n[i]%10;}c.len=i;while(c.n[c.len+1]>0){c.len++;c.n[c.len+1]=c.n[c.len]/10;c.n[c.len]=c.n[c.len]%10;}return c;}Bigint inttimes(Bigint a,int b){Bigint c;memset(b.n,0,sizeof(b.n));c.digit=a.digit;int i;for(i=1;i<=a.len;i++){c.n[i]=}}int compare(Bigint a,Bigint b){int i,j,k;        if(a.len>b.len)    {    return 1;    }    else if(a.len<b.len)    {    return -1;    }    else    {    for(i=a.len;i>=1;i--)        {        if(a.n[i]>b.n[i])                return 1;            else if(a.n[i]<b.n[i])                return -1;        }    }    return 0;}
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