題意：輸入一個進制b（[2,20]），確定從[1,300]中哪些數的平方在該進制下是回文數

解法：核心部分是將十進制數轉換為任意進制數，除x取余，逆序排列

/* ID: lsswxr1 PROG: palsquare LANG: C++ */ #include <iostream> #include <vector> #include <map> #include <list> #include <set> #include <deque> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <cctype> #include <cstdio> #include <iomanip> #include <cmath> #include <cstdio> #include <string> #include <fstream> using namespace std;///宏定義 const int INF = 1000000000; const int MAXN = 15; const int maxn = MAXN; ///全局變量 和 函數#define USACO #ifdef USACO #define cin fin #define cout fout #endif // int b; stack<int> stk, origin, square; queue<int> que; char str[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; int main() {#ifdef USACO ofstream fout ("palsquare.out");ifstream fin ("palsquare.in"); #endif ///變量定義while(cin >> b){for (int i = 1; i <= 300; i++){bool flag = true;while (!origin.empty())origin.pop();while (!square.empty())square.pop();while (!stk.empty())stk.pop();while (!que.empty())que.pop();int k = i * i;int o = i;while (o != 0){int yuu = o % b;origin.push(yuu);o /= b;}while (k != 0){int yu = k % b;stk.push(yu);que.push(yu);square.push(yu);k /= b;}while ((!stk.empty()) && (!que.empty())){int f1, f2;f1 = stk.top();f2 = que.front();if (f1 != f2){flag = false;break;}stk.pop();que.pop();}if (flag){while (!origin.empty()){int index = origin.top();cout << str[index];origin.pop();}cout << " ";while (!square.empty()){int index = square.top();cout << str[index];square.pop();}cout << endl;}}}///結束return 0; }

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官方的解法：

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #include <ctype.h> #include <math.h>/* is string s a palindrome? */ int ispal(char *s) {char *t;t = s+strlen(s)-1;for(t=s+strlen(s)-1; s<t; s++, t--)if(*s != *t)return 0;return 1; }/* put the base b representation of n into s: 0 is represented by "" */ // 進制轉換的遞歸寫法，可以借鑒和學習 void numbconv(char *s, int n, int b) {int len;if(n == 0) {strcpy(s, "");return;}/* figure out first n-1 digits */numbconv(s, n/b, b);/* add last digit */len = strlen(s);s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b];s[len+1] = '\0'; }void main(void) {char s[20];char t[20];int i, base;FILE *fin, *fout;fin = fopen("palsquare.in", "r");fout = fopen("palsquare.out", "w");assert(fin != NULL && fout != NULL);fscanf(fin, "%d", &base);for(i=1; i <= 300; i++) {numbconv(s, i*i, base);if(ispal(s)) {numbconv(t, i, base);fprintf(fout, "%s %s\n", t, s);}}exit(0); }

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轉載于:https://www.cnblogs.com/rayforsure/p/3440947.html

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