一堆的头文件和宏不用看,看solve函数
我用了一个set来判断是不是每行都只有一个,最后判断数量即可
然后由于对角线的性质,一条是相加和相同,一条是相减差相同,判断即可
时间复杂度O(n^2),再加上K的200
C++ 代码
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define ppb pop_back
#define lbnd lower_bound
#define ubnd upper_bound
#define endl '\n'
#define trav(a, x) for(auto& a : x)
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (ll)x.size()
#define hell 1000000007
#define DEBUG cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x) cerr << #x << " is " << x << endl;
#define ini(a) memset(a,0,sizeof(a))
#define ini2(a,b) memset(a,b,sizeof(a))
#define case ll T;read(T);for(ll Q=1;Q<=T;Q++)
#define lowbit(x) x&(-x)
#define pr printf
#define sc scanf
#define _ 0
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \
(void)(cout << "L" << __LINE__ \
<< ": " << #x << " = " << (x) << '\n')
#define TIE \
cin.tie(0);cout.tie(0);\
ios::sync_with_stdio(false);
//#define long long int
//using namespace __gnu_pbds;
template <typename T>
void read(T &x) {
x = 0;
int f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') f = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + (ch ^ 48);
ch = getchar();
}
x *= f;
return;
}
inline void write(long long x) {
if(x<0) putchar('-'), x=-x;
if(x>9) write(x/10);
putchar(x%10+'0');
putchar('\n');
}
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 50009;
const ll N = 5;
int arr[1099];
set<int> si;
void solve(){
si.clear();
int q, flag = 0;
cin>>q;
for (int i=1; i<=q; i++) {
cin>>arr[i];
si.insert(arr[i]);
}
if (si.size() != q) {
cout<<"NO"<<endl;
return ;
}
for (int i=1; i<=q && flag==0; i++) {
int p = i - arr[i], o = i + arr[i];
for (int j=1; j<=q && flag==0; j++) {
if (j == i) continue;
if (j - arr[j] == p || j + arr[j] == o) {
cout<<"NO"<<endl;
return ;
}
}
}
cout<<"YES"<<endl;
}
int main()
{
// TIE;
#ifndef ONLINE_JUDGE
// freopen ("in.txt" , "r", stdin );
// freopen ("out.txt", "w", stdout);
#else
#endif
// solve();
case{solve();}
// case{cout<<"Case "<<Q<<":"<<endl;solve();}
}