sgu491

求满足ax + by = n,(x >= 1, y >= 1)的a,b有多少个。

比赛时候没有做出来，是因为想到只知道n,要求a,b,x,y枚举的话4个铁定超时

如这类找满足条件个数的，如果不能暴力枚举的话都要尽量去转化成为求因子个数。

转化成n - ax = by，枚举a,x,那么b的个数就相当于n - ax中的因子个数，依次检查每个因子，要比a大，然后去重统计。

#include <cstdio>#include <string.h>#include <vector>#include <algorithm>using namespace std;const int MAX_NUMBER = 100005;vector<int> factor[MAX_NUMBER];int temp_ans[MAX_NUMBER * 50];int temp_factor[MAX_NUMBER];int number;void getAllFactor(int n) {int factor_number = 0;for (int i = 1; i * i <= n; i++) {if (n % i == 0) {if (i * i == n) {temp_factor[factor_number++] = i;}else {temp_factor[factor_number++] = i;temp_factor[factor_number++] = n / i;}}}sort(temp_factor, temp_factor + factor_number);for (int i = 0; i < factor_number; i++) {factor[n].push_back(temp_factor[i]);}}int main() {for (int i = 2; i < MAX_NUMBER; i++) {getAllFactor(i);}while (scanf("%d", &number) != EOF) {long long ans = 0;for (int a = 1; a <= number / 2; a++) {int temp_number = 0;for (int x = 1; number - a * x > 0; x++) {int by = number - a * x;for (int i = 0; i < factor[by].size(); i++) {int b = factor[by][i];if (b > a) {temp_ans[temp_number++] = b;}}}sort(temp_ans, temp_ans + temp_number);if (temp_number > 0) {ans++;}for (int i = 1; i < temp_number; i++) {if (temp_ans[i] != temp_ans[i - 1]) {ans++;}}}printf("%I64d\n", ans);}return 0;}