HDUOJ 1002 A+B problem Ⅱ
题意简述要点分析参考代码
题目来源
题意简述
[INPUT] The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. [OUTPUT] For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
输入一个数T(1<=T<=20)表示有T个测试用例，接下来T行每行输入a和b两个数。输入的a和b均为正整数，但数字的长度不超过1000。给出a+b的和。
要点分析
本质就是大数加法的运用
严格的格式要求 1.输入为一次输入，输出也为一次输出。 2.每个输出结果之间有一行空行。 3.最后一个输出结果后有且仅有一个空行。
参考代码
#include<iostream>
#include<string>
#include<vector>
using namespace std
;
int main()
{
string a
,b
;
vector
<string
> res
[3];
int t
,r
;
cin
>>t
;
while(t
--)
{
cin
>>a
>>b
;
res
[0].push_back(a
);
res
[1].push_back(b
);
if(a
.size()<b
.size())
{
r
=b
.size()-a
.size();
for(int i
=a
.size()-1;i
>=0;i
--)
{
b
[r
+i
]+=a
[i
]-'0';
if(b
[r
+i
]>'9')
{
b
[r
+i
]-=10;
b
[r
+i
-1]++;
}
}
for(int i
=a
.size()-1;i
>=0;i
--)
{
if(b
[i
]>'9')
{
b
[i
]-=10;
b
[i
-1]++;
}
}
if(b
[0]>'9')
{
b
[0]-=10;
b
.insert(0,1,'1');
}
res
[2].push_back(b
);
}
else
{
r
=a
.size()-b
.size();
for(int i
=b
.size()-1;i
>=0;i
--)
{
a
[r
+i
]+=b
[i
]-'0';
if(a
[r
+i
]>'9' && r
+i
!=0)
{
a
[r
+i
]-=10;
a
[r
+i
-1]++;
}
}
for(int i
=b
.size()-1;i
>=0;i
--)
{
if(a
[i
]>'9' && i
!=0)
{
a
[i
]-=10;
a
[i
-1]++;
}
}
if(a
[0]>'9')
{
a
[0]-=10;
a
.insert(0,1,'1');
}
res
[2].push_back(a
);
}
}
for(int i
=0;i
<res
[0].size();i
++)
{
if(i
!=0)cout
<<endl
;
cout
<<"Case "<<i
+1<<":"<<endl
;
cout
<<res
[0][i
]<<" + "<<res
[1][i
]<<" = "<<res
[2][i
]<<endl
;
}
return 0;
}
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