A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 310717 Accepted Submission(s):
60078
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
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
Recommend
解题心得：
这是个大数问题，题意要求计算不超过1000位的两个数字相加，所以必须要用字符串存储两个数字。
我刚开始选择的是让他们从最低位开始相加，但是两个字符串需要考虑长度问题，写完后有发现很多问题。
忽略了进位问题，可能不止进一位，当99999+1时就需要进好几位。结果一遍一遍的改就是不对，下面的代码是错误的。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int n; char a[1005],b[1005],ab[1005]; int c_a,c_b; int i1=0,i2=0,ii; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%s %s",a,b); printf("Case %d: ",i+1); c_a=strlen(a); c_b=strlen(b); if(c_a<c_b){ for(int j=c_a;j>=0;j--){ a[j+c_b-c_a+1]=a[j]; } for(int j=c_b-1;j>=0;j--){ b[j+1]=b[j]; } memset(a,'0',(c_b-c_a+1)*sizeof(char)); memset(ab,'0',c_b*sizeof(char)); for(int j=c_a+1;j>=0;j--){ ii=(int)(a[j]-48)+(int)(b[j]-48)+i2; i1=ii/10; i2=ii%10; ab[j]=(char)(i2+48); i2=i1; } ab[c_b+1]=' '; if(i2!=0){ ab[0]=(char)(i2+48); } if(ab[0]=='0'){ printf("%s + ",&a[c_b-c_a+1]); printf("%s = ",b+1); printf("%s ",ab+1); }else{ printf("%s + ",&a[c_b-c_a+1]); printf("%s = ",b+1); printf("%s ",ab); } i2=i1=0; } if(c_a==c_b){ for(int j=c_a-1;j>=0;j--){ a[j+1]=a[j]; } for(int j=c_a-1;j>=0;j--){ b[j+1]=b[j]; } memset(a,'0',1*sizeof(char)); memset(b,'0',1*sizeof(char)); memset(ab,'0',1*sizeof(char)); for(int j=c_a;j>=0;j--){ ii=(int)(a[j]-48)+(int)(b[j]-48)+i2; i1=ii/10; i2=ii%10; ab[j]=(char)(i2+48); i2=i1; } ab[c_a+1]=' '; if(ab[0]!='0'){ printf("%s + ",a+1); printf("%s = ",b+1); printf("%s ",ab); } else{ printf("%s + ",a+1); printf("%s = ",a+1); printf("%s ",&ab[1]); } i1=i2=0; } if(c_a>c_b){ for(int j=c_b;j>=0;j--){ b[j+c_a-c_b]=b[j]; } memset(b,'0',(c_a-c_b)*sizeof(char)); for(int j=c_b;j>=0;j--){ ii=(int)(a[j]-48)+(int)(b[j]-48)+i2; i1=ii/10; i2=ii%10; ab[j]=(char)(i2+48); i2=i1; } i1=i2=0; ab[c_a]=' '; printf("%s + ",&a[0]); printf("%s = ",&b[c_a-c_b]); printf("%s ",&ab[0]); } } return 0; }
然后上网百度之后发现反着计算更简单，把低位存储在啊a[0]的位置，高位往后存，这样不用考虑长度问题，进位问题也好解决。
下面是代码：
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int n; int i1=0; char a[1000],b[1000]; int la,lb; int c[1000]={0}; scanf("%d",&n); for(int t=1;t<=n;t++){ scanf("%s %s",&a,&b); la=strlen(a)-1; lb=strlen(b)-1; for(int i=la;i>=0;i--){ c[i1++]=(a[i]-'0'); } i1=0; for(int i=lb;i>=0;i--){ c[i1++]+=(b[i]-'0'); if(c[i1-1]>=10){ c[i1-1]-=10; c[i1]++; } } while(c[i1]!=0){ if(c[i1]>=10){ c[i1]-=10; c[i1+1]++; } i1++; } printf("Case %d: ",t); printf("%s + %s = ",a,b); i1=la>lb?la:lb; if(c[i1+1]!=0){ printf("%d",c[i1+1]); } for(int j=i1;j>=0;j--){ printf("%d",c[j]); } i1=0; if(t!=n){ printf(" "); }else{ printf(" "); } memset(c,0,1000*sizeof(int)); } return 0; }
这个题也要注意格式问题！！
又重新做了这个题，下面是代码：
#include <iostream> #include <cstdio> #include <cstring> using namespace std; char a[1001]={'0'}; char b[1001]={'0'}; int c[1001]={0}; int main() { int g,s; int n; cin>>n; for(int cou=0;cou<n;cou++){ scanf("%s",&a[1]); scanf("%s",&b[1]); int la=strlen(a); int lb=strlen(b); if(la<lb){ for(int i=la-1,j=lb-1,k=lb-1;i>0&&j>0;i--,j--,k--){ c[k]+=(int)(a[i]-'0')+(int)(b[j]-'0'); if(c[k]>=10){ g=c[k]%10; s=c[k]/10; c[k]=g; c[k-1]+=s; } } for(int i=lb-la;i>=0;i--){ c[i]+=(int)(b[i]-'0'); if(c[i]>=10){ g=c[i]%10; s=c[i]/10; c[i]=g; c[i-1]+=s; } } } if(la>lb){ for(int i=lb-1,j=la-1,k=la-1;i>0&&j>0;i--,j--,k--){ c[k]+=(int)(b[i]-'0')+(int)(a[j]-'0'); if(c[k]>=10){ g=c[k]%10; s=c[k]/10; c[k]=g; c[k-1]+=s; } } for(int i=la-lb;i>=0;i--){ c[i]+=(int)(a[i]-'0'); if(c[i]>=10){ g=c[i]%10; s=c[i]/10; c[i]=g; c[i-1]+=s; } } } if(la==lb){ for(int i=lb-1,j=la-1;i>0&&j>0;i--,j--){ c[i]+=(int)(b[i]-'0')+(int)(a[j]-'0'); if(c[i]>=10){ g=c[i]%10; s=c[i]/10; c[i]=g; c[i-1]+=s; } } } int l= la>=lb ? la:lb; printf("Case %d: ",cou+1); printf("%s + ",&a[1]); printf("%s = ",&b[1]); if(c[0]!=0){ for(int i=0;i<l;i++){ printf("%d",c[i]); } }else{ for(int i=1;i<l;i++){ printf("%d",c[i]); } } memset(c,0,1001*sizeof(int)); //if(cou!=n-1){ printf(" "); //} } return 0; }