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
解析:高精度
参考代码:
//HDU1002 高精度 #include<iostream> #include<string> #include<string.h> using namespace std; string a,b,c,d; void solve(){ int res[1005]; int i ,j,k,r; i = j = k = r = 0; bool flag = false; memset(res,0,sizeof(res)); cin>>a>>b; c = a; d = b; if(a.size()<b.size()||(a.size()==b.size()&&a<b)){//把把大数放前面. swap(a,b); } for(i = a.size()-1,j = b.size() - 1; j>=0;i--,j--){ res[k++] = (r+a[i]-'0'+b[j]-'0') % 10; r = (r+a[i]-'0'+b[j]-'0') / 10;//记录进位 } while(i>=0){//再把大数的其他部分进行运算 res[k++] = (r+a[i]-'0') % 10; r = (r+a[i]-'0') / 10; i--; } if(r){ res[k++] = r;//如果还有进位,进到最高位 } cout<<c<<" "<<"+"<<" "<<d<<" "<<"="<<" "; for(i = k - 1; i >=0;i--){ if(res[i]||flag){ printf("%d",res[i]); flag = true; } } if(!flag){ cout<<0; } cout<<endl; } int main() { int cas; cin>>cas; for(int i = 0;i < cas;i++){ if(i){ cout<<endl; } printf("Case %d:\n",i+1); solve(); } return 0; } ############################################### #include<iostream> #include<algorithm> #include<string.h> using namespace std; int main() { char a[1005],b[1005]; int i,j,n,s1[1005],s2[1005],len1,len2,len,cnt,res[1005]; bool flag; scanf("%d",&n); for(cnt = 0;cnt < n;cnt++){ if(cnt){ cout<<endl; } flag = false; memset(s1,0,1005*sizeof(int)); memset(s2,0,1005*sizeof(int)); memset(res,0,1005*sizeof(int)); memset(a,'0',sizeof(a)); memset(b,'0',sizeof(b)); scanf("%s",a); scanf("%s",b); len1 = strlen(a); len2 = strlen(b); len = max(len1,len2); //cout<<len<<endl; j = 0; for(i = len1 - 1; i >=0;i--){ s1[j++] = a[i]- '0' ;//倒叙存到 s1中 } j = 0; for(i = len2 - 1;i>=0;i--){ s2[j++] = b[i]- '0'; } for(i = 0;i < len;i++){ res[i] += s1[i]+s2[i]; if(res[i]>=10){ res[i+1]++; res[i] %= 10; } // s1[i]+=s2[i]; // if(s1[i]>=10){ // s1[i]-=10; // s1[i+1]++; // } } printf("Case %d:\n",cnt+1); printf("%s + %s = ",a,b); for(i = 1005;i>=0;i--){ if(res[i]!=0||flag){ cout<<res[i]; flag = true; } } if(!flag){ cout<<0; } cout<<endl; } }
