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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#855996#9738. Make It Divisibleicpc_zhzx034#WA 1ms4504kbC++142.2kb2025-01-13 14:29:132025-01-13 14:29:14

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  • [2025-01-13 14:29:14]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:4504kb
  • [2025-01-13 14:29:13]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
#define fi first
#define se second
#define mkp make_pair
#define pb emplace_back
#define popcnt __builtin_popcountll
const ll mod = 998244353;
inline ll read(){
	ll x=0, f=1; char ch=getchar();
	while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
	while(ch>='0' && ch<='9') x=x*10+ch-'0', ch=getchar();
	return x*f;
}
inline int lg2(int x){ return 31^__builtin_clz(x); }
inline ll lg2(ll x){ return 63^__builtin_clzll(x); }
inline void addmod(int &x){ if(x >= mod) x -= mod; }
inline void addmod(ll &x){ if(x >= mod) x -= mod; }
inline ll qpow(ll a,ll b){
	ll ans=1, base=a;
	while(b){
		if(b&1) ans=ans*base%mod;
		base=base*base%mod; b>>=1;
	}
	return ans;
}
inline ll INV(ll x){ return qpow(x, mod-2); };

unordered_map<ll,ll>mp;

ll cnt,p[50005],q[50005];
ll n,k,b[50005],lc[50005],rc[50005],stk[50005],tp;
void procedure(){
	mp.clear();	memset(lc,0,sizeof(lc)); memset(rc,0,sizeof(rc));
	tp=cnt=0;
	n=read(),k=read();
	for(ll i=1;i<=n;i++) b[i]=read();

	for(ll i=1;i<=n;i++){
		while(tp && b[stk[tp]]>b[i]){
			lc[i]=stk[tp];
			tp--;
		}
		if(tp) rc[stk[tp]]=i;
		stk[++tp]=i;
	}

	for(ll i=1;i<=n;i++){
		if(lc[i] && b[lc[i]] != b[i]){
			p[++cnt] = b[i], q[cnt] = b[lc[i]];
		}
		if(rc[i] && b[rc[i]] != b[i]){
			p[++cnt] = b[i], q[cnt] = b[rc[i]];
		}
	}
	if(!cnt){
		printf("%lld %lld\n", k, k*(k-1)/2);
		return;
	}

	// p[1]+d | q[1]-p[1]
	int x = q[1]-p[1];

	vector<int>v;
	ll ans=0, sum=0;
	for(int d=1;d*d<=x;d++){
		if(x%d) continue;
		if(d>p[1] && d-p[1]<=k) v.pb(d-p[1]);
		if(d*d==x) continue;
		if((x/d)>p[1] && (x/d)-p[1]<=k) v.pb((x/d)-p[1]);
	}


	for(auto d: v){
		bool flg=0;
		for(int i=1;i<=n;i++){
			if((q[i]+d)%(p[i]+d)){
				flg = 1;
				break;
			}
		}
		if(!flg){
			ans ++;
			sum += d;
		}
	}
	printf("%lld %lld\n", ans, sum);
}
int main(){
	#ifdef LOCAL
		assert(freopen("input.txt","r",stdin));
		assert(freopen("output.txt","w",stdout));
	#endif
	ll T=read();
	// math_init();
	// NTT::init();
	while(T--) procedure();
	return 0;
}

Details

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Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 4504kb

input:

3
5 10
7 79 1 7 1
2 1000000000
1 2
1 100
1000000000

output:

3 8
0 0
100 4950

result:

wrong answer 3rd lines differ - expected: '100 5050', found: '100 4950'