poj2756解题报告

Autumn is a Genius

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 6040	Accepted: 1269

Description

Jiajia and Wind have a very cute daughter called Autumn. She is so clever that she can do integer additions when she was just 2 years old! Since a lot of people suspect that Autumn may make mistakes, please write a program to prove that Autumn is a real genius.

Input

The first line contains a single integer T, the number of test cases. The following lines contain 2 integers A, B(A, B < 32768) each. The size of input will not exceed 50K.

Output

The output should contain T lines, each with a single integer, representing the corresponding sum.

Sample Input

1
1 2
      

Sample Output

3      

Hint

There may be '+' before the non-negative number! 题目大意：Autumn是一个天才，他能计算A+B的和 (A, B < 32768)....... 思路：这个题玩文字游戏，只给定了AB的上限而没给定AB下线，靠的。。。Autumn真是个天才，哎无奈 乖乖的用高精加减吧。。。汗 #include<iostream>

using namespace std;

char a[100000],b[100000],c[1000000];int lena,lenb,lenc;

void plus()

{

int jin=0,i;

lena=strlen(a)-1;

lenb=strlen(b)-1;

if(lena>lenb)

{ strcpy(c,a); lenc=lena; }

else

{ strcpy(c,b); strcpy(b,a); lenc=lenb; lenb=lena; }

while(lenb>-1)

{

i=c[lenc]+b[lenb]+jin-96;

c[lenc]=i%10+48;

jin=i/10;

lenc--; lenb--;

}

while(lenc>-1&&jin)

{

i=c[lenc]-48+jin;

c[lenc]=i%10+48;

jin=i/10;

lenc--;

}

if(jin)

{

for(i=strlen(c)+1;i>0;i--)

c[i]=c[i-1];

c[0]=jin+48;

}

}

void jian()

{

int jie=0,i,j;

lena=strlen(a)-1;

lenb=strlen(b)-1;

lenc=lena;

strcpy(c,a);

while(lenb>-1)

{

if(c[lenc]-b[lenb]-jie<0)

{

c[lenc]=c[lenc]-b[lenb]-jie+58;

jie=1;

}

else

{ c[lenc]=c[lenc]-b[lenb]-jie+48; jie=0; }

lenb--;lenc--;

}

while(lenc>-1&&jie)

{

if(c[lenc]-48-jie<0)

{ c[lenc]='9'; jie=1; }

else

{ c[lenc]-=jie; jie=0;}

lenc--;

}

for(i=0;c[i]=='0'&&i<strlen(c)-1;i++);

for(j=0;;j++)

{

c[j]=c[i+j];

if(c[j]=='/0')

break;

}

}

int main()

{

int t,i;

cin>>t;

while(t--)

{

int f1=1,f2=1;

cin>>a;

if(a[0]=='-'||a[0]=='+')

{

if(a[0]=='-')

f1=-1;

for(i=0;a[i]!='/0';i++)

a[i]=a[i+1];

}

cin>>b;

if(b[0]=='-'||b[0]=='+')

{

if(b[0]=='-')

f2=-1;

for(i=0;b[i]!='/0';i++)

b[i]=b[i+1];

}

if(f1*f2==1)

{

plus();

if(f1==-1)

cout<<'-'<<c<<endl;

else

cout<<c<<endl;

}

else if(f1*f2==-1)

{

if(f1==1)

{

if(strlen(a)>strlen(b)||(strlen(a)==strlen(b)&&strcmp(a,b)>0))

{

jian();

cout<<c<<endl;

}

else

{

strcpy(c,a); strcpy(a,b); strcpy(b,c);

jian();

if(strcmp(c,"0"))

cout<<'-'<<c<<endl;

else

cout<<"0"<<endl;

}

}

else

{

if(strlen(a)>strlen(b)||(strlen(a)==strlen(b)&&strcmp(a,b)>0))

{

jian();

if(strcmp(c,"0"))

cout<<'-'<<c<<endl;

else

cout<<"0"<<endl;

}

else

{

strcpy(c,a); strcpy(a,b); strcpy(b,c);

jian();

cout<<c<<endl;

}

}

}

}

return 0;

}