前面的两个就不写了。以后我没做完一个例题就会发表一篇。不过时间不定，我有时也很少有时间来练习这些题。
/* acm_1002 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 */ #include<iostream> #include<cstring> using namespace std; char *add(char *a,char *b) //定义函数 { int i,j,k=0; //循环变量的定义 int temp[1001]; int length1 = strlen(a); int length2 = strlen(b); for(i = length1-1, j = length2 -1 ; i >= 0&&j >= 0; --i,--j) temp[k++] = a[i] + b[j] - '0' - '0'; for(;i >= 0; --i) temp[k++] = a[i] - '0'; for(;j >= 0; --j) temp[k++] = b[j] - '0'; temp[k] = 0; for(i = 0;i < k; ++i) { temp[i+1] += temp[i] / 10; temp[i] %=10; } if(!temp[k]) k--; for(i = 0;i <= k;i++) a[i] = temp[k-i] + '0'; a[k+1] = '/0'; return a; } int main() { char a[1001],b[1001],*r; int number = 0; int k = 1; cin >> number; while(number--) { if(cin >> a >> b) { cout << "Case "<< k++ << ":/n" << a << " + " << b <<" = "; r = add(a,b); cout << r; cout << endl; if(number) cout<<endl; } } return 0; }
