C语言大数四则运算

/**/////////////////
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <ctype.h> 
/*int cchkdig(char *r) 
{ 
    int i=0;  
    while(r[i]!=‘\0‘) 
    { 
        if(isdigit(r[i++])==0) 
            return (0); 
    } 
    return (1);  
} */
void del_0(char *r)				//去掉整数串表示前面多余的零，最后结果为空串时置为"0" 
{
    unsigned int lr;  
    int i=0, j; 
    lr=strlen(r);  
    while(r[i]==‘0‘) 
        ++i;
    if(i>0) 
    {
        for(j=0; j<lr-i; ++j)
            r[j]=r[j+i];
        for(j=lr-i; j<lr; ++j) 
        {
            r[j]=‘\0‘; 
        }
    }
    if(r[0]==‘\0‘)
    {
        r[0]=‘0‘;
    } 
} 
int scmp(char *r, char *u)			//比较长度
{ 
    unsigned int lr, lu; 
    del_0(r);
    del_0(u); 
    lr=strlen(r); 
    lu=strlen(u); 
    if(lr>lu) 
    {
        return 1;
    }
    else if (lr<lu) 
    {
        return -1;
    }
    return (strcmp(r, u)); 
}
char *sub(char *r, char *u)					//两个串表示数的减法 
{
    unsigned int i,lr, lu, lp,c=0;  
    char h,hc; 
    char *p;
    if(scmp(r, u)<0)						//r比u大
        return NULL; 
    lr=strlen(r); 
    lu=strlen(u); 
    p=(char *)malloc((unsigned int)(lr+1)*sizeof(char)); 
    for(i=0; i<lu; ++i) 
    {
        h=r[lr-i-1]-u[lu-i-1]-c; 
        if(h<0) 
        {
            c=1; 
            h=h+10;
        } 
        else 
            c=0; 
        p[i]=h+‘0‘;
        hc=h+‘0‘;
    } 
    for (i=lu; i<lr; ++i) 
    { 
        h=r[lr-i-1]-‘0‘-c;
        if(h<0) 
        {
            c=1; 
            h=h+10; 
        } 
        else 
            c=0; 
        p[i]=‘0‘+h; 
        hc=‘0‘+h;
    }
    p[i]=‘\0‘;
    lp=i-1;
    while(p[lp]==‘0‘&&lp!=0)
    {
        p[lp]=‘\0‘;
        lp--; 
    }
    for(i=0; i<(lp+1)/2; ++i) 
    { 
        hc=p[i]; 
        p[i]=p[lp-i]; 
        p[lp-i]=hc; 
    } 
    return (p); 
}//end sub() 
char *add(char *r, char *u)			//两个串表示数的加法 
{ 
    unsigned int lr, lu, lp; 
    int i, h, c=0; 
    char hc, *p; 
    lr=strlen(r); 
    lu=strlen(u); 
    if(lu>lr) 
    { 
        p=r; 
        r=u; 
        u=p; 
        h=lr; 
        lr=lu; 
        lu=h; 
    } 
    p=(char *)malloc((unsigned int)(lr+2)*sizeof(char));  
    for(i=0; i<lu; ++i) 
    { 
        h=r[lr-i-1]-‘0‘+u[lu-i-1]-‘0‘+c; 
        if(h>9) 
        { 
            c=1;
            h=h-10; 
        } 
        else 
            c=0;  
        p[i]=h+‘0‘;  
    } 
    for(i=lu; i<lr; ++i) 
    { 
        h=r[lr-i-1]-‘0‘+c; 
        if(h>9) 
        { 
            c=1; 
            h=h-10; 
        } 
        else 
            c=0;  
        p[i]=‘0‘+h;  
    } 
    if(c>0) 
    { 
        p[i]=c+‘0‘; 
        lp=i; 
    } 
    else 
        lp=i-1;  
    for(i=lp+1; i<lr+2; ++i) 
        p[i]=‘\0‘; 
    for(i=0; i<(lp+1)/2; ++i) 
    { 
        hc=p[i]; 
        p[i]=p[lp-i]; 
        p[lp-i]=hc; 
    } 
    return (p); 
}//end add() 
char *mul(char *r, char *u)			//两个串表示数的乘法 
{ 
    unsigned int lr, lu, lp; 
    int i, j, c, h; 
    char *p; 
    lr=strlen(r); 
    lu=strlen(u); 
    p=(char *)malloc((unsigned int)(lr+lu+1)*sizeof(char)); 
    for(i=0; i<lr+lu; ++i) 
        p[i]=‘0‘; 
    p[lr+lu]=‘\0‘; 
    for(i=lr-1; i>=0; --i) 
    { 
        c=0;  
        for(j=lu-1; j>=0; --j) 
        { 
            lp=i+j+1;  
            h=(r[i]-‘0‘)*(u[j]-‘0‘)+p[lp]-‘0‘+c;  
            c=h/10;  
            h=h%10;  
            p[lp]=h+‘0‘;  
        } 
        if(c>0)p[i+j+1]=c+‘0‘;  
    } 
    del_0(p);  
    return p;  
}//end mul() 
char *div(char *u, char *v, int n)					//两个串表示数的除法，结果精确到小数点后第n位 
{ 
    char *p, *f, *r,*q;  
    unsigned int i, lu, lv, lr,  iw, c, h;  
    int  kh, j;  
    lu=strlen(u);  
    lv=strlen(v);  
    f=(char *)malloc((unsigned int)(lu+n+3)*sizeof(char)); 
    q=(char *)malloc(sizeof(char)); 
    for(i=0; i<lu+n+3; ++i) 
        f[i]=‘\0‘; 
    r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); 
    for(i=0; i<lv+2; ++i) 
        r[i]=‘\0‘; 
    for(iw=0; iw<lu+n+2; ++iw) 
    { 
        if(iw<lu) 
        { 
            del_0(r); 
            lr=strlen(r); 
            r[lr]=u[iw]; 
            r[lr+1]=‘\0‘;
        } 
        else if(iw>lu) 
        { 
            del_0(r);
            q[0]=‘0‘;
            if(scmp(r, q)==0)
            {
                break; 
            }
            lr=strlen(r); 
            r[lr]=‘0‘;
            r[lr+1]=‘\0‘; 
        } 
        else 
        { 
            f[lu]=‘.‘; 
            continue;  
        }
        kh=0; 
        while(scmp(r, v)>=0) 
        { 
            p=r;
            r=sub(p, v); 
            ++kh; 
        } 
        f[iw]=kh+‘0‘; 
    } 
    if(iw==lu+n+2) 
    { 
        if(f[lu+n+1]>=‘5‘) 
        { 
            f[lu+n+1]=‘\0‘;  
            c=1;  
            for(j=lu+n; j>=0; --j) 
            { 
                if(c==0) 
                {
                    break;  
                }
                if(f[j]==‘.‘)
                {
                    continue; 
                }
                h=f[j]-‘0‘+c;  
                if(h>9) 
                { 
                    h=h-10; 
                    c=1; 
                } 
                else 
                    c=‘\0‘;  
                f[j]=h+‘0‘; 
            } 
        } 
        else 
            f[lu+n+1]=‘\0‘; 
    } 
    free(r); 
    free(p);
    q=NULL;
    free(q);
    del_0(f);  
    return(f);  
}//end div() 
/*
//两个串表示数的除法，结果分别用整商与余数表示 
char *sdivkr(char *u, char *v, char **rout) 
{ 
    char  *f, *r;  
    unsigned int i, lu, lv, lr, iw;  
    int  kh;  
    lu=strlen(u);  
    lv=strlen(v);  
    f=(char *)malloc((unsigned int)(lu+1)*sizeof(char)); 
    for(i=0; i<lu+1; ++i) f[i]=‘\0‘; 
    r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); 
    for(i=0; i<lv+2; ++i) r[i]=‘\0‘; 
    for(iw=0; iw<lu; ++iw) 
    { 
        del_0(r); 
        lr=strlen(r); 
        r[lr]=u[iw]; 
        r[lr+1]=‘\0‘; 
        kh=0; 
        while(scmp(r, v)>=0) 
        { 
            r=sub(r, v); 
            ++kh; 
        } 
        f[iw]=kh+‘0‘; 
    } 
    del_0(r);  
    *rout=r;  
    del_0(f); 
    return(f);
}//end *sdivkr() 
*/
/*调用上述函数实现两任意长正整数任意指定精度的算术计算器程序 
void main(int argc, char *argv[]) 
{ 
    char *p, *r;  
    int n;
	 if(argc!=4)
    { 
        if(argc!=3)
            printf("\n>>\"order n1 op n2\" or n ! "); 
        exit(0); 
    } 
    del_0(argv[1]);  
    if(cchkdig(argv[1])==0)
    { 
        printf("Input data error, Input again!"); 
        exit(0); 
    } 
    del_0(argv[3]);  
    if(cchkdig(argv[3])==0) 
    { 
        printf("Input data error, Input again!"); 
        exit(0); 
    } 
    if(strcmp(argv[2], "+")==0) 
    { 
        printf("%s", p=add(argv[1], argv[3])); 
        free(p); 
    } 
    else if(strcmp(argv[2], "-")==0) 
    { 
        printf("%s", p=sub(argv[1], argv[3])); 
        free(p); 
    } 
    else if(strcmp(argv[2], "*")==0) 
    { 
        printf("%s", p=mul(argv[1], argv[3])); 
        free(p); 
    } 
    else if(argv[2][0]==‘/‘ && strlen(argv[2])==1) 
    { 
        if(argv[3][0]==‘0‘)
        {
            printf("error!devided by zero!!\n");
            exit(0);
        }
        p=sdivkr(argv[1], argv[3], &r);  
        printf("k=%s r=%s", p, r);  
        free(p); 
        free(r);  
    } 
    else 
    if(argv[2][0]==‘/‘&&strlen(argv[2])>1) 
    { 
        if(argv[3][0]==‘0‘)
        {
            printf("error!devided by zero!!\n");
            exit(0);
        }
        argv[2][0]=‘\0‘;  
        del_0(argv[2]);  
        if(cchkdig(argv[2])==0) 
        { 
            printf("Input data error, Input again!"); 
            exit (0); 
        } 
        n=atoi(argv[2]);  
        printf("%s", p=div(argv[1], argv[3], n));
        free(p);  
    } 
} */
char a[200000];
int  main()
{
	//freopen("read.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	while(n--)
	{
		memset(a, ‘\0‘, sizeof(a));
		char b[200] = {"0"};
		while(~scanf("%s", a) && a[0] != 0)
		{
			strcpy(b, add(a, b));
		}
		printf("%s\n", b);
		if(n!=0)	printf("\n");
	}
	return 0;
}

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