题解 | 高精度整数加法
高精度整数加法
https://www.nowcoder.com/practice/49e772ab08994a96980f9618892e55b6
#include <iostream> #include <string> using namespace std; #define N 10010 int a[N]; int b[N]; int res[N]; void transfer(string& s1, string& s2){ // 将s2 和 s2 转换成一个个数字并存储在a , b 数组中 int len1 = s1.size(); int len2 = s2.size(); for(int i = 0; i < len1; ++i) a[len1 - i - 1] = s1[i] - '0'; for(int i = 0; i < len2; ++i) b[len2 - i - 1] = s2[i] - '0'; } void Add(){ int carry = 0; // 进位位 for(int i = 0; i < N; ++i){ res[i] = (a[i] + b[i] + carry) % 10; carry = (a[i] + b[i] + carry) / 10; } } int main() { string str1; string str2; cin >> str1; cin >> str2; if(str1 == "0" && str2 == "0"){ cout << 0 << endl; return 0; } transfer(str1,str2); Add(); int i = N - 5; while(res[i] == 0) --i; while(i >= 0) printf("%d",res[i--]); cout << endl; return 0; }