/**/////////////////
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
/*int cchkdig(char *r)
{
int i=0;
while(r[i]!=' ')
{
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]=' ';
}
}
if(r[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]=' ';
lp=i-1;
while(p[lp]=='0'&&lp!=0)
{
p[lp]=' ';
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]=' ';
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]=' ';
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]=' ';
r=(char *)malloc((unsigned int)(lv+2)*sizeof(char));
for(i=0; i<lv+2; ++i)
r[i]=' ';
for(iw=0; iw<lu+n+2; ++iw)
{
if(iw<lu)
{
del_0(r);
lr=strlen(r);
r[lr]=u[iw];
r[lr+1]=' ';
}
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]=' ';
}
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]=' ';
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=' ';
f[j]=h+'0';
}
}
else
f[lu+n+1]=' ';
}
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]=' ';
r=(char *)malloc((unsigned int)(lv+2)*sizeof(char));
for(i=0; i<lv+2; ++i) r[i]=' ';
for(iw=0; iw<lu; ++iw)
{
del_0(r);
lr=strlen(r);
r[lr]=u[iw];
r[lr+1]=' ';
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("
>>"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!!
");
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!!
");
exit(0);
}
argv[2][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, ' ', sizeof(a));
char b[200] = {"0"};
while(~scanf("%s", a) && a[0] != 0)
{
strcpy(b, add(a, b));
}
printf("%s
", b);
if(n!=0) printf("
");
}
return 0;
}