USACO1.2.3 Name That Number (namenum)

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扫一遍dict，再扫每一位。

不过如果当前扫到的名字的长度和输入数字的长度不同就可以跳过了

/*ID:shijiey1PROG:namenumLANG:C++*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;string dict[5000];string num;char letter[10][3] = {{},{}, {'A','B','C'}, {'D','E','F'}, {'G','H','I'}, {'J','K','L'}, {'M','N','O'}, {'P','R','S'}, {'T','U','V'}, {'W','X','Y'}};int n = 0;bool contains(char c, int i) {return letter[i][0] == c || letter[i][1] == c || letter[i][2] == c;}int main() {freopen("namenum.out", "w", stdout);freopen("namenum.in", "r", stdin);cin >> num;freopen("dict.txt", "r", stdin);while (cin >> dict[n++]);bool has = false;for (int i = 0; i < n; i++) {if (dict[i].size() != num.size()) continue;bool flag = false;for (int j = 0; j < num.size(); j++) {if (!contains(dict[i][j], num[j] - '0')) {flag = true;break;}}if (!flag) {cout << dict[i] << endl;has = true;}}if (!has) {printf("NONE\n");}return 0;}

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