题目大意

题目链接

给你两个数n(<= 5000), m(<= 1e9), n为你要构造的序列长度, m下面会用到。该序列满足

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1. 序列递增
2. 1 <= ai <= 1e9
3. 有恰好m组 i, j, k (1 <= i < j < k <= n) 满足 ai + aj = ak

如果存在这样的序列, 输出这个序列, 不存在输出-1、

分析

转自此

1

代码

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/*
power by Solo_Dance
*/
#include <bits/stdc++.h>
#define eps 1e-8
using namespace std;
#define ms(a, b) memset((a), (b), sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const ll ll_max = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

inline ll read() {
ll res = 0;bool f = 0;char ch = getchar();
while (ch < '0' || ch > '9') {if (ch == '-') f = 1;ch = getchar();}
while (ch <= '9' && ch >= '0') {res = (res << 3) + (res << 1) + ch - '0';ch = getchar();}
return f ? (~res + 1) : res;
}
int main(){
int n = read(), m = read();
int k = 0, sum = 0;
while(k < n && sum + k / 2 <= m) sum += k / 2, k++;
if (k == n && sum < m) return puts("-1"), 0;
for (int i = 1; i <= k; ++i) printf("%d ", i);
if (k == n) return 0;
printf("%d ", k - 2 * (m - sum) + 1 + k);
for (int i = k + 2; i <= n; ++i) printf("%d ", int(1e8 + (i - k - 2) * 1e4));

return 0;
}
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