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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#767916 | #9738. Make It Divisible | i_wish_a_girl_friend# | WA | 1ms | 7952kb | C++23 | 3.3kb | 2024-11-20 22:42:37 | 2024-11-20 22:42:37 |
Judging History
answer
#include<bits/stdc++.h>
#define int long long
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define pb push_back
#define pii pair<int,int>
using namespace std;
//const int mod=998244353;
int qpw(int a,int b,int mod){int ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
//int inv(int x){return qpw(x,mod-2);}
const int MAX_N = 5e4+10;
const int LOG = 17; // log2(MAX_N) 的上界
int rs_min[MAX_N],ls_min[MAX_N],root_min;
int n, a[MAX_N];
int up[MAX_N][LOG]; // 倍增表:up[i][k] 表示从 i 向上跳 2^k 步后的节点
int gcd_table[MAX_N][LOG]; // gcd_table[i][k] 表示从 i 到 up[i][k] 的 GCD 值
int parent[MAX_N]; // 笛卡尔树中的父节点
int depth[MAX_N]; // 每个节点的深度
int query_min(int l,int r){
int x=root_min;
for(;;x=(r<x?ls_min[x]:rs_min[x]))if(l<=x&&x<=r)return x;
}
int st[17][MAX_N + 1]; // ST 表
int logs[MAX_N + 1]; // log 预处理数组
// 求 GCD
int gcd(int a, int b) {
return b?gcd(b,a%b):a;
}
// ST 表预处理
void init() {
for (int i = 1; i < 17; ++i) {
for (int j = 1; j + (1 << i) - 1 <= n; ++j) {
st[i][j] = gcd(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
}
}
}
// 区间查询
int query(int l, int r) {
int k = logs[r - l + 1];
return gcd(st[k][l], st[k][r - (1 << k) + 1]);
}
void solve(){
int k;cin>>n>>k;
set<int>sb;
for(int i=1,x;i<=n;i++)cin>>a[i],sb.insert(a[i]);
if(sb.size()==1){
cout<<k<<" "<<k*(k+1)/2<<'\n';
return;
}
vector<int>sta(n+1);
function<void()>build_min=[&](){
int top=0,cur=0;
for(int i=1;i<=n;i++){
cur=top;
while(cur&&a[sta[cur]]>=a[i])cur--;
if(cur)rs_min[sta[cur]]=i;
if(cur<top)ls_min[i]=sta[cur+1];
sta[++cur]=i;
top=cur;
}
};
build_min();
root_min=sta[1];
vector<int>c(n+1);
for(int i=1;i<=n;i++)c[i]=a[i];
int gc=0,mx=0;
for(int i=n;i>=2;i--)a[i]=abs(a[i]-a[i-1]),gc=gcd(gc,a[i]);
for(int i=n;i>=2;i--)st[0][i]=a[i];
st[0][1]=0;
init();
set<int>dp,ndp;
function<void(int,int)>dfs=[&](int l,int r){
if(l>=r)return;
if(l==1&&r==n){
int G=query(l+1,r);
int mid=query_min(l,r);
for(int i=1;i*i<=G;i++){
if(G%i)continue;
if(i-c[mid]>0&&i-c[mid]<=k)
dp.insert(i-c[mid]);
if(G/i-c[mid]>0&&G/i-c[mid]<=k)
dp.insert(G/i-c[mid]);
}
dfs(l,mid-1);
dfs(mid+1,r);
}
else {
ndp.clear();
int G=query(l+1,r);
int mid=query_min(l,r);
for(int x:dp){
if(G%(c[mid]+x))continue;
ndp.insert(x);
}
swap(ndp,dp);
dfs(l,mid-1);
dfs(mid+1,r);
}
};
dfs(1,n);
int sum=0;
for(int x:dp)sum+=x;
cout<<dp.size()<<" "<<sum<<'\n';
}
signed main(){
ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
int _=1;
cin>>_;
while(_--)solve();
}
/*
3
5 10
7 79 1 7 1
2 1000000000
1 2
1 100
1000000000
*/
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 5880kb
input:
3 5 10 7 79 1 7 1 2 1000000000 1 2 1 100 1000000000
output:
3 8 0 0 100 5050
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 0ms
memory: 7952kb
input:
4 201 1000000000 1 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5...
output:
0 0 0 0 0 0 0 0
result:
ok 4 lines
Test #3:
score: -100
Wrong Answer
time: 1ms
memory: 7672kb
input:
500 4 1000000000 8 14 24 18 4 1000000000 17 10 18 14 4 1000000000 6 17 19 19 4 1000000000 15 14 15 25 4 1000000000 16 16 5 25 4 1000000000 4 30 20 5 4 1000000000 11 4 23 9 4 1000000000 14 25 13 2 4 1000000000 18 18 1 15 4 1000000000 22 22 22 28 4 1000000000 15 17 17 10 4 1000000000 22 14 13 25 4 100...
output:
0 0 0 0 0 0 0 0 2 20 0 0 0 0 0 0 3 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
result:
wrong answer 5th lines differ - expected: '0 0', found: '2 20'