都是上面那个编程作业 我的思路是用栈进行逆序执行 然后就能不用递归做出来了
#include<iostream>
#include<stack> #include<string> #include<cctype> #include<sstream> #include <strstream> #define l_num 22 using namespace std; bool isNum(string str) { string num = "1234567890"; if (str.find_first_not_of(num) == string::npos) return true; return false; } string inttow(int n) { string str = ""; while (n) { string temp(1, n % 10 + '0'); str = temp + str; n -= n % 10; n =int(n/10.0); } return str; } int main() { int num; int i; cin >> num; string s[20];//本来是动态分配的 string s1; int n, x; string b; for (i = 0; i < num; i++) { cin >> s[i]; } do { cin >> b; if (b == "over") { break; } else {//如果没有结束 string a[l_num]; a[0] = b; i = 1; if (a[0] == "printall") { int cal; for (cal = 0; cal < num; cal++) { cout << s[cal]; cout << endl; } } else { do { cin >> a[i]; i++; } while (getchar()!='\n'); stack<int> st; stack<string> ss; for (i--; i >= 0; i--) { int temp; if (isdigit(a[i][0]))//如果是数字 进行转换 并压栈 { temp = atoi(a[i].c_str());//进行数字转换 并压栈 st.push(temp); } else { if (a[i] == "find") { x = st.top(); st.pop(); n = st.top();st.pop(); stringstream sd; sd << x; string str = sd.str(); int k = s[n - 1].find(str);//改了 0 k = (k == string::npos ? s[n - 1].length() : k); st.push(k);//执行find 的功能 } else if (a[i] == "rfind") { int s2, n; s2 = st.top(); st.pop(); n = st.top(); st.pop(); stringstream sd; sd << s2; string str = sd.str(); int k = s[n - 1].rfind(str); k = (k == string::npos ? s[n - 1].length() : k); st.push(k);//执行find 的功 } else if (a[i] == "add") { string s2; string c; s1 = ss.top(); ss.pop(); s2= ss.top(); ss.pop(); if (isNum(s1) && isNum(s2)) { int a = atoi(s1.c_str()); int b = atoi(s2.c_str()); char aa[100]; //10是十进制的意思 //openjudge上面，不能使用itoa和sprintf // itoa(a+b, aa, 10); // sprintf(aa, "%d", a+b); strstream ss; int sum = a + b; ss << sum; ss >> c; } else c = s1 + s2; ss.push(c);//执行find 的功 } else if (a[i] == "copy") { string t; int n,x, l; n = st.top(); st.pop(); x = st.top(); st.pop(); l = st.top(); st.pop(); t = s[n - 1].substr(x, l); ss.push(t);//执行find 的功 } else if (a[i] == "reset") {//将第n个字符串变为s string s2 = ss.top(); ss.pop(); int n= st.top(); st.pop(); s[n - 1] = s2; } else if (a[i] == "insert") { s1 = ss.top(); ss.top(); int n, x; n = st.top(); st.pop(); x= st.top(); st.pop(); s[n - 1].insert(x, s1); } else if (a[i] == "print") { int n = st.top(); st.pop(); cout << s[n - 1]; cout << endl; } else { ss.push(a[i]); } } } } } } while (b != "over"); return 0; }