题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 315232 Accepted Submission(s): 61142
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
Author
Ignatius.L
题解:大数A+B,直接用模板来解决(附上不同的代码)
#include <iostream> #include <cstring> using namespace std; const int M=1000+10; string add(string a,string b) { string c; int len1=a.length(); int len2=b.length(); int len=max(len1,len2); for(int i=len1;i<len;i++) a="0"+a; for(int i=len2;i<len;i++) b="0"+b; int ok=0; for(int i=len-1;i>=0;i--) { char temp=a[i]+b[i]-'0'+ok; if(temp>'9') { ok=1; temp-=10; } else ok=0; c=temp+c; } if(ok) c="1"+c; return c; } int main() { int t; string a,b; cin>>t; for(int i=1;i<=t;i++) { if(i>1) cout<<endl; cin>>a>>b; cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "<<add(a,b)<<endl; } return 0; }
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; const int M=100000+10; string add(char * a,char * b) { char d[M]="\0",e[M]="\0"; int len1=strlen(a); int len2=strlen(b); int len=max(len1,len2); d[0]=e[0]='0'; for(int i=len1,j=1;i<len;i++) d[j++]='0'; strcat(d,a); for(int i=len2,j=1;i<len;i++) e[j++]='0'; strcat(e,b); //cout<<"d="<<d<<",e="<<e<<endl; len=strlen(d);//cout<<"len="<<len<<endl; for(int i=len-1;i>=0;i--) { int x=d[i]-'0',y=e[i]-'0'; if(x+y>9) { d[i-1]+=1; d[i]+=y-10; } else {d[i]+=y;} } if (d[0]=='0')return (d+1); else return d; } int main() { int t; char a[M],b[M]; cin>>t; for(int i=1;i<=t;i++) { if(i>1) printf("\n"); scanf("%s%s",a,b); cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "<<add(a,b)<<endl; } return 0; }
#include<cstdio> #include<cstring> using namespace std; int main() { int j=1,p=0,i,n,aa,bb; char a[1000],b[1000],c[1000]; scanf("%d",&n); while(n) { scanf("%s%s",a,b); printf("Case %d:\n",j++); printf("%s + %s = ",a,b); aa=strlen(a)-1;bb=strlen(b)-1; for( i=0;aa>=0||bb>=0;aa--,bb--,i++) { if(aa>=0&&bb>=0)c[i]=a[aa]+b[bb]-'0'+p; if(aa>=0&&bb<0)c[i]=a[aa]+p; if(aa<0&&bb>=0)c[i]=b[bb]+p; if(c[i]>'9'){c[i]=c[i]-10;p=1;} else p=0; } if(p==1)printf("%d",p); while(i--)printf("%c",c[i]); if(n==1) printf("\n"); else printf("\n\n"); n--; } return 0; }
版权声明:此代码归属博主, 请务侵权!