A + B Problem (大数相加3种方法) (转自孙悦学长)
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Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
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.
Sample Input
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
解题思路:把两个数的每一位都存放到一个数组中,逆序存放,然后对应位相加的和存放到另一个数组中。
方法1:
- #include <iostream>
- #include <string.h>
- using namespace std;
- int main()
- {
- char x[1001],y[1001];//用来存放输入的两个数,第一个大数存在x里面,第二个大数存在y里面
- int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1;//xx数组(整型)用来存放x数组中数字的逆序,因为做加法的时候要从原数的末位开始相加,sum数组用来存放两个大数每位对应相加的值
- int lenx,leny,maxlen,t;
- cin>>t;
- while(t--)
- {
- cin>>x>>y;
- lenx=strlen(x);//x的长度
- leny=strlen(y);//y的长度
- maxlen=lenx>leny?lenx:leny;//把最大长度赋给maxlen
- for(i=0;i<lenx;i++)
- xx[i]=x[lenx-i-1]-'0';//根据ascii码表 '0'为48例如 '4'为52 ,相减得到4 ,存放到整型数组xx中,注意x[lenx-i-1],意思是说逆序存放,为下面逆序相加做准备
- for(i=0;i<leny;i++)
- yy[i]=y[leny-i-1]-'0';
- for(i=0;i<maxlen;i++)
- {
- sum[i]+=xx[i]+yy[i];//对应位相加
- sum[i+1]=sum[i]/10;//因为两个一位数相加不会超过18,如果sum[i]大于10,则sum[i+1]为1,也就是向高位进1
- sum[i]=sum[i]%10;//如果对应为相加大于10,取余,相当于 -10;
- }
- if(sum[i]==1)
- maxlen++;//要注意i的值,它代表最高位的后一位,如果最高位大于10大话,要进1,也就是判断s[i]是否为一,进一意味着让位数maxlen+1
- cout<<"Case "<<j<<":"<<endl;
- j++;
- cout<<x<<" + "<<y<<" = ";
- for(i=maxlen-1;i>=0;i--)
- cout<<sum[i];
- cout<<endl;
- if(t!=0)
- cout<<endl;//没有这一句,AC不了,格式问题
- for(i=0;i<maxlen;i++)
- {
- sum[i]=0;
- xx[i]=0;
- yy[i]=0;
- }//重置,为下一次输入做准备
- }
- }
#include <iostream>#include <string.h>using namespace std;int main(){ char x[1001],y[1001];//用来存放输入的两个数,第一个大数存在x里面,第二个大数存在y里面 int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1;//xx数组(整型)用来存放x数组中数字的逆序,因为做加法的时候要从原数的末位开始相加,sum数组用来存放两个大数每位对应相加的值 int lenx,leny,maxlen,t; cin>>t; while(t--) { cin>>x>>y; lenx=strlen(x);//x的长度 leny=strlen(y);//y的长度 maxlen=lenx>leny?lenx:leny;//把最大长度赋给maxlen for(i=0;i<lenx;i++) xx[i]=x[lenx-i-1]-'0';//根据ascii码表 '0'为48例如 '4'为52 ,相减得到4 ,存放到整型数组xx中,注意x[lenx-i-1],意思是说逆序存放,为下面逆序相加做准备 for(i=0;i<leny;i++) yy[i]=y[leny-i-1]-'0'; for(i=0;i<maxlen;i++) { sum[i]+=xx[i]+yy[i];//对应位相加 sum[i+1]=sum[i]/10;//因为两个一位数相加不会超过18,如果sum[i]大于10,则sum[i+1]为1,也就是向高位进1 sum[i]=sum[i]%10;//如果对应为相加大于10,取余,相当于 -10; } if(sum[i]==1) maxlen++;//要注意i的值,它代表最高位的后一位,如果最高位大于10大话,要进1,也就是判断s[i]是否为一,进一意味着让位数maxlen+1 cout<<"Case "<<j<<":"<<endl; j++; cout<<x<<" + "<<y<<" = "; for(i=maxlen-1;i>=0;i--) cout<<sum[i]; cout<<endl; if(t!=0) cout<<endl;//没有这一句,AC不了,格式问题 for(i=0;i<maxlen;i++) { sum[i]=0; xx[i]=0; yy[i]=0; }//重置,为下一次输入做准备 }}
方法2:
- #include <iostream>
- #include <string.h>
- using namespace std;
- int main()
- {
- char x[1001],y[10001];
- int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
- int t;
- cin>>t;
- while(t--)
- {
- cin>>x>>y;
- lenx=strlen(x);
- leny=strlen(y);
- maxlen=lenx>leny?lenx:leny;
- for(i=0;i<lenx;i++)
- xx[i]=x[lenx-i-1]-'0';
- for(i=0;i<leny;i++)
- yy[i]=y[leny-i-1]-'0';
- for(i=0;i<maxlen;i++)
- {
- sum[i]+=xx[i]+yy[i];
- sum[i+1]=sum[i]/10;
- sum[i]=sum[i]%10;//算法和方法1相同
- }
- if(sum[i]==1)
- maxlen++;
- cout<<"Case "<<j<<":"<<endl;
- j++;
- cout<<x<<" + "<<y<<" = ";
- for(i=maxlen-1;i>=0;i--)
- cout<<sum[i];
- cout<<endl;
- if(t!=0)
- cout<<endl;
- memset(xx,0,1001*sizeof(int));//,把xx数组里的值清0,memset是按照字节进行赋值,一开始写成memset(xx,0,1001);错误,整型四个字节,需要1001*4,即1001*sizeof(int)
- memset(yy,0,1001*sizeof(int));
- memset(sum,0,1001*sizeof(int));
- }
- }
#include <iostream>#include <string.h>using namespace std;int main(){ char x[1001],y[10001]; int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen; int t; cin>>t; while(t--) { cin>>x>>y; lenx=strlen(x); leny=strlen(y); maxlen=lenx>leny?lenx:leny; for(i=0;i<lenx;i++) xx[i]=x[lenx-i-1]-'0'; for(i=0;i<leny;i++) yy[i]=y[leny-i-1]-'0'; for(i=0;i<maxlen;i++) { sum[i]+=xx[i]+yy[i]; sum[i+1]=sum[i]/10; sum[i]=sum[i]%10;//算法和方法1相同 } if(sum[i]==1) maxlen++; cout<<"Case "<<j<<":"<<endl; j++; cout<<x<<" + "<<y<<" = "; for(i=maxlen-1;i>=0;i--) cout<<sum[i]; cout<<endl; if(t!=0) cout<<endl; memset(xx,0,1001*sizeof(int));//,把xx数组里的值清0,memset是按照字节进行赋值,一开始写成memset(xx,0,1001);错误,整型四个字节,需要1001*4,即1001*sizeof(int) memset(yy,0,1001*sizeof(int)); memset(sum,0,1001*sizeof(int)); }}
方法3:
- #include <iostream>
- #include <string.h>
- using namespace std;
- int main()
- {
- char x[1001],y[10001];
- int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
- int t;
- cin>>t;
- while(t--)
- {
- cin>>x>>y;
- lenx=strlen(x);
- leny=strlen(y);
- maxlen=lenx>leny?lenx:leny;
- for(i=0;i<lenx;i++)
- xx[i]=x[lenx-i-1]-'0';
- for(i=0;i<leny;i++)
- yy[i]=y[leny-i-1]-'0';
- for(i=0;i<maxlen;i++)
- {
- sum[i]+=xx[i]+yy[i];
- if(sum[i]>=10)//判断两个大数的对应位相加是否大于等于10
- {
- sum[i+1]=1;//如果大于10,高位进1
- sum[i]-=10;//如果大于20,本位-10
- }
- }
- if(sum[i]==1)
- maxlen++;
- cout<<"Case "<<j<<":"<<endl;
- j++;
- cout<<x<<" + "<<y<<" = ";
- for(i=maxlen-1;i>=0;i--)
- cout<<sum[i];
- cout<<endl;
- if(t!=0)
- cout<<endl;
- memset(xx,0,1001*sizeof(int));
- memset(yy,0,1001*sizeof(int));
- memset(sum,0,1001*sizeof(int));
- }
- }
#include <iostream>#include <string.h>using namespace std;int main(){ char x[1001],y[10001]; int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen; int t; cin>>t; while(t--) { cin>>x>>y; lenx=strlen(x); leny=strlen(y); maxlen=lenx>leny?lenx:leny; for(i=0;i<lenx;i++) xx[i]=x[lenx-i-1]-'0'; for(i=0;i<leny;i++) yy[i]=y[leny-i-1]-'0'; for(i=0;i<maxlen;i++) { sum[i]+=xx[i]+yy[i]; if(sum[i]>=10)//判断两个大数的对应位相加是否大于等于10 { sum[i+1]=1;//如果大于10,高位进1 sum[i]-=10;//如果大于20,本位-10 } } if(sum[i]==1) maxlen++; cout<<"Case "<<j<<":"<<endl; j++; cout<<x<<" + "<<y<<" = "; for(i=maxlen-1;i>=0;i--) cout<<sum[i]; cout<<endl; if(t!=0) cout<<endl; memset(xx,0,1001*sizeof(int)); memset(yy,0,1001*sizeof(int)); memset(sum,0,1001*sizeof(int)); }}
运行截图:
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