C. Everyone is a Winner!
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
On the well-known testing system MathForces, a draw of nn rating units is arranged. The rating will be distributed according to the following algorithm: if kk participants take part in this event, then the nn rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.
For example, if n=5n=5 and k=3k=3, then each participant will recieve an 11 rating unit, and also 22 rating units will remain unused. If n=5n=5, and k=6k=6, then none of the participants will increase their rating.
Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.
For example, if n=5n=5, then the answer is equal to the sequence 0,1,2,50,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋⌊n/k⌋ for some positive integer kk (where ⌊x⌋⌊x⌋ is the value of xx rounded down): 0=⌊5/7⌋0=⌊5/7⌋, 1=⌊5/5⌋1=⌊5/5⌋, 2=⌊5/2⌋2=⌊5/2⌋, 5=⌊5/1⌋5=⌊5/1⌋.
Write a program that, for a given nn, finds a sequence of all possible rating increments.
Input
The first line contains integer number tt (1≤t≤101≤t≤10) — the number of test cases in the input. Then tt test cases follow.
Each line contains an integer nn (1≤n≤1091≤n≤109) — the total number of the rating units being drawn.
Output
Output the answers for each of tt test cases. Each answer should be contained in two lines.
In the first line print a single integer mm — the number of different rating increment values that Vasya can get.
In the following line print mm integers in ascending order — the values of possible rating increments.
Example
input
Copy
4 5 11 1 3
output
Copy
4 0 1 2 5 6 0 1 2 3 5 11 2 0 1 3 0 1 3
题意:
就是问你 n / (1 -- INF) 有多少种答案
思路:1.分块数论
2.找规律
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main() {
int t;
scanf("%d", &t);
while(t--) {
int n;
scanf("%d", &n);
vector<int>A;
map<int,int>B;
A.push_back(0);
A.push_back(1);
for(int i = n>>1;i >= 1;--i) {
int h = n/i;
if(B[h] == 0) {
A.push_back(h);
B[h] = 1;
}
}
printf("%d\n", A.size());
for(int i = 0;i < A.size();++i) {
printf("%d ", A[i]);
}
putchar('\n');
}
return 0;
}
//暴力代码找规律
ac代码:
思路:推出1-sqrt(n)都是答案然后后面的暴力查找即可
#include<bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while(t--) {
int n;vector<int>A;map<int,int>B;
scanf("%d", &n);
int g = sqrt(n);
for(int i = 0;i <= g;++i) {
A.push_back(i);
B[i] = 1;
}
for(int i = sqrt(n);i >= 1;--i) {
int h = n/i;
if(B[h] == 0) {
A.push_back(h);
B[h] = 1;
}
}
printf("%d\n", A.size());
for(int i = 0;i < A.size();++i) {
printf("%d ", A[i]);
}
putchar('\n');
}
return 0;
}
数论分块板子过:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
int ans[MAXN];
int main(){
int T;
cin >> T;
while(T--){
int n;
scanf("%d",&n);
int cnt = 0;
ans[cnt++] = 0;
for(int l = 1, r = 0; l <= n; l++) {
r = n / (n / l);
l = r;
if(r != ans[cnt - 1]){
ans[cnt++] = r;
}
}
printf("%d\n",cnt);
for(int i = 0; i < cnt; i++){
printf("%d ",ans[i]);
}
puts("");
}
}