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思路：这题与csu1392题目类似，方法类似。枚举最高位，最低位和中间数字的长度，然后列等式，计算中间的数字，看长度是不是跟枚举的一致，需要注意的是中间数字可以有前导0，如果根据等式算出来的中间数字为K，枚举的长度为L，也就是说需要满足length(K)<=L。
csu1392: http://www.cnblogs.com/jklongint/p/4419007.html
代码：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 | #pragma comment(linker, "/STACK:10240000,10240000")#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <algorithm>#include <queue>using namespace std;long long ans[1234], MI[12], X;int cnt;int length(long long x) { int ans = 0; while (x) { ans ++; x /= 10; } return ans;}void solve(long long L, long long a, long long b) { long long ga = (a * X - b * MI[6]) * MI[L + 1] + b * X - a * MI[6]; long long gb = MI[7] - X * 10, K = ga / gb; if (ga % gb == 0 && K >= 0 && length(K) <= L) ans[cnt ++] = a * MI[L + 1] + K * 10 + b;}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE MI[0] = 1; for (int i = 1; i < 12; i ++) MI[i] = MI[i - 1] * 10; int T, cas = 0; cin >> T; while (T --) { printf("Case #%d:\n", ++ cas); double x; cin >> x; X = (long long)(x * MI[6] + 0.1); if (X == MI[6]) { cout << 0 << endl; continue; } cnt = 0; for (int L = 0; L < 9; L ++) { for (int a = 1; a < 10; a ++) { for (int b = 0; b < 10; b ++) { solve(L, a, b); } } } cout << cnt << endl; for (int i = 0; i < cnt; i ++) { printf("%I64d%c", ans[i], i == cnt - 1? ‘\n‘ : ‘ ‘); } } return 0;} |
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原文地址：http://www.cnblogs.com/jklongint/p/4579007.html
