Satisfactory Pairs HackerRank - pairs-again——预处理＋不定长数组应用

Think:

1题意：输入n, 求(a, b)满足a < b 且存在x, y使得x*a + y*b = n的二元组个数，要求n, a, b, x, y皆为正整数

2方法：先预处理小于n的数的约数，然后暴力试探＋剪枝，试探符合题意的二元组个数，时间复杂度 1/x 从1到n积分，为nln(n)

vjudge题目链接

建议参考博客１

建议参考博客２

以下为Accepted代码

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 4;

vector <int> y_b
;
vector <int> :: iterator b;

int book
;

int main(){
int n, i, j, a, x, t, ans;
scanf("%d", &n);
for(i = 1; i <= n; i++){
for(j = 1; j*j <= i; j++){
if(i%j == 0){
y_b[i].push_back(j);
if(i*i != j){
t = i / j;
y_b[i].push_back(t);
}
}
}
sort(y_b[i].begin(), y_b[i].end(), greater<int>());
}
ans = 0;
for(a = 1; a < n; a++){
for(x = 1; x*a < n; x++){
t = n - x*a;
for(b = y_b[t].begin(); b != y_b[t].end(); b++){
if((*b) <= a)
break;
if(book[*b] != a){
ans++;
book[*b] = a;
}
}
}
}
printf("%d\n", ans);
return 0;
}

以下为建议参考代码

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 4;

vector <int> y_b
;/*记录当前元素的约数*/
vector <int> :: iterator b;

int book
;/*实数型全局变量初始化默认为0*/

bool cmp(int a, int b){
return a > b;
}

int main(){
int n, i, j, a, x, t, ans;
scanf("%d", &n);
for(i = 1; i <= n; i++){
for(j = 1; j*j <= i; j++){
if(i%j == 0){
y_b[i].push_back(j);
if(i*i != j){
t = i / j;
y_b[i].push_back(t);
}
}
}
sort(y_b[i].begin(), y_b[i].end(), greater<int>());
}
ans = 0;
for(a = 1; a < n; a++){
for(x = 1; x*a < n; x++){
t = n - x*a;
for(b = y_b[t].begin(); b != y_b[t].end(); b++){
if((*b) <= a)/*题意：a < b*/
break;
if(book[*b] != a){/*标记判重*/
ans++;
book[*b] = a;
}
}
}
}
printf("%d\n", ans);
return 0;
}