codeforces 233c Cycles【贪心】

Cycles

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

John Doe started thinking about graphs. After some thought he decided that he wants to paint an undirected graph, containing exactly kcycles
of length 3.

A cycle of length 3 is an unordered group of three distinct graph vertices a, b and c,
such that each pair of them is connected by a graph edge.

John has been painting for long, but he has not been a success. Help him find such graph. Note that the number of vertices there shouldn't exceed 100,
or else John will have problems painting it.

Input

A single line contains an integer k (1 ≤ k ≤ 105)
— the number of cycles of length 3 in the required graph.

Output

In the first line print integer n (3 ≤ n ≤ 100)
— the number of vertices in the found graph. In each of next n lines print n characters
"0" and "1": the i-th
character of the j-th line should equal "0",
if vertices i and j do
not have an edge between them, otherwise it should equal "1". Note that as the required graph is undirected, the i-th
character of the j-th line must equal the j-th
character of the i-th line. The graph shouldn't contain self-loops, so the i-th
character of the i-th line must equal "0"
for all i.

Examples

input

1

output

3
011101110

input

10

output

5
0111110111110111110111110

这道题坑了我好几个小时，首先算循环节小于k的全满的状态，在一个节点然后一条边的加上去。例如，K = 5，当K=4的时候是全满的状态，此时只需要添加一个节点和边，使其再产生一个环。通过规律可以知道，因为添加一个节点以前是全满的状态，所以对于一个新增加的节点，第一次添加两条边，干掉一个环，以后每增加一条边，会干掉2、3、4……用计数器记录一下就好。

如果全填满了还不满足,在增加一个节点如此进行，必定可以保证最优解。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 100 + 10;
int ans[104][104];
int main() {
int K, temp, t;
while (scanf("%d", &K) != EOF) {
memset(ans, 0, sizeof(ans));
for (int i = 3; i <= 100; i++) {
if (i*(i - 1)*(i - 2)/6 >= K) {
t = i;
temp = i*(i - 1)*(i - 2)/6;
break;
}
}
if (temp == K) {
for (int i = 0; i < t; i++) {
for (int j = 0; j < t; j++) {
ans[i][j] = !(i == j);
}
}
}
else {
K -= (t - 1)*(t - 2)*(t - 3)/6;
for (int i = 0; i < t - 1; i++) {
for (int j = 0; j < t - 1; j++) {
ans[i][j] = !(i == j);
}
}
for (; t <= 100; t++) {
int cnt = 1;
while (K >= cnt) {
if (cnt == 1) {
ans[t - 1][0] = ans[t - 1][1] = 1;
ans[0][t - 1] = ans[1][t - 1] = 1;
}
else {
for (int i = 0; i < t; i++) {
bool flag = false;
for (int j = 0; j < t; j++) {
if (!ans[i][j] && i != j) {
ans[i][j] = ans[j][i] = 1;
flag = true; break;
}
}
if (flag) break;
}
}
K -= cnt; cnt++;
}
if (K == 0) break;
}
}
printf("%d\n", t);
for (int i = 0; i < t; i++) {
for (int j = 0; j < t; j++) {
printf("%d", ans[i][j]);
}
printf("\n");
}
}
return 0;
}