PS:换了一种方式 希望大家喜欢 2333
/** code by: zstu wxk time: 2019/03/01 Problem Link: http://codeforces.com/contest/1130/problem/E solve: 如果构造的数列是 0 0 0 0 0 0 0 -1 + + + +(+代表大于0的数) 那么 find_answer = sum(+) * len_+ 实际上的答案为 tot_len * [sum(+) - 1] 即: sum(+) * len_+ + k = tot_len * [sum(+) - 1] 接下来就是枚举tot_len就好了 **/ #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 1e5 + 100; int n, k, t1, t2; int a[N]; bool check(int x){ for(int i = 2; i * i <= x; ++i){ if(x % i == 0){ t1 = i, t2 = x/i; t1 = n - t1; if(t1 < n - 2 && t2 < 1e6 * t1) return true; } } return false; } void Ac(){ n = 2000; memset(a, 0, sizeof a); for(n = 2000; n >= 1; --n){ if(check(n+k)){ for(int i = 1; i <= t1; ++i){ if(t2 >= 1e6){ a[i] = 1e6; t2 -= 1e6; } else{ a[i] = t2; t2 = 0; } } a[t1+1] = -1; printf("%d ", n); reverse(a+1, a+1+n); for(int i = 1; i <= n; ++i){ printf("%d ", a[i]); } break; } } } int main(){ while(~scanf("%d", &k)){ Ac(); } return 0; }