Codeforces Round #827 (Div. 4)

A

水题

#include<bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;void solve() {int a,b,c;cin>>a>>b>>c;if(a+b==c||a+c==b||b+c==a)cout<<"YES"<<endl;else cout<<"NO"<<endl;}
signed main(){int t=1;cin>>t;while (t--) solve();return 0;
}

B

水题

#include<bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;void solve() {int n;cin>>n;vector<int> a(n);for(int i=0;i<n;i++) cin>>a[i];sort(a.begin(),a.end());for(int i=0;i<n-1;i++){if(a[i]==a[i+1]){cout<<"NO"<<endl;return;}}cout<<"YES"<<endl;}
signed main(){int t=1;cin>>t;while (t--) solve();return 0;
}

C

如果有一整行或者一整列为同一个字母，说明肯定是最后一次画的。

#include<bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;char g[20][20];void solve() {for(int i=0;i<8;i++){for(int j=0;j<8;j++){cin>>g[i][j];}}for(int i=0;i<8;i++){int cnt=0;for(int j=0;j<8;j++){if(g[i][j]=='R') cnt++;}if(cnt==8){cout<<"R"<<endl;return;}}for(int i=0;i<8;i++){int cnt=0;for(int j=0;j<8;j++){if(g[j][i]=='B') cnt++;}if(cnt==8){cout<<"B"<<endl;return;}}
}
signed main(){int t=1;cin>>t;while (t--) solve();return 0;
}

D

桶计数，暴力找,两个数互质，gcd为1

#include<bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;void solve() {int n;cin>>n;vector<int> a(n+1);vector<int> pos(1001,-1);for(int i=1;i<=n;i++) {cin>>a[i];pos[a[i]]=i;}int ans=0;int ok=0;for(int i=1;i<=1000;i++){for(int j=1;j<=1000;j++){if(__gcd(i,j)==1&&pos[i]!=-1&&pos[j]!=-1){ans=max(ans,pos[i]+pos[j]);ok=1;}}}if(!ok) cout<<-1<<endl;else cout<<ans<<endl;}
signed main(){int t=1;cin>>t;while (t--) solve();return 0;
}

E

预处理，前缀和，二分

#include <bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
#define int long long
void solve()
{int n, q;cin >> n >> q;vector<int> a(n);for (int i = 0; i < n; i++)cin >> a[i];vector<int> sum(n);sum[0] = a[0];for (int i = 1; i < n; i++) {sum[i] = sum[i - 1] + a[i];}for (int i = 1; i < n; i++) {a[i] = max(a[i], a[i - 1]);}// for (auto& x : a)//     cout << x << " ";// cout << endl;// for (auto& x : sum)//     cout << x << " ";// cout << endl;while (q--) {int x;cin >> x;int l = 0, r = n - 1;int ans = 0;while (l <= r) {int mid = (l + r) / 2;if (x >= a[mid]) {ans = sum[mid];l = mid + 1;} else {r = mid - 1;}}cout << ans << " ";}cout << endl;
}
signed main()
{int t = 1;cin >> t;while (t--)solve();return 0;
}

F

注意开始的时候S,T字符串都是有一个a的，可以发现如果T字符串存在比a字符大，我们就可以让S的a字符放在第一个，T的大于a的字符放在第一个，这样肯定S<T。

如果T全是a没有比a大的，就看S有没有比a大的，如果有，则无法构造。如果没有，S也完全是a字符，则比较a字符的数量。

#include <bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
#define int ll
void solve()
{int q;cin >> q;int hs = 0, ht = 0, cs = 0, ct = 0;while (q--) {int op, k;string x;cin >> op >> k >> x;if (op == 1) {for (auto& c : x) {if (c > 'a') {hs = 1;} else {cs += k;}}} else {for (auto& c : x) {if (c > 'a') {ht = 1;} else {ct += k;}}}// cout << hs << " " << ht << endl;if (ht) {cout << "YES" << endl;} else if (!hs) {if (cs < ct)cout << "YES" << endl;elsecout << "NO" << endl;} else if (hs) {cout << "NO" << endl;}}
}
signed main()
{int t = 1;cin >> t;while (t--)solve();return 0;
}
/*
s:aabb
t:aaaaa
*/

G

观察可以发现，第一个数肯定是a数组中最大的那个数，那么接下来就找和前面或最大的数，可以遍历一遍没有用过的数，找到最大的那个，由于最多有31位，所以最多遍历31次或的结果就是最大的,遍历31次之后，剩下就可以随便输出了。

#include<bits/stdc++.h>
#define sc(x) scanf("%lld", &(x))
#define pr(x) printf("%lld\n", (x))
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;void solve() {int n;cin>>n;vector<int> a(n);for(int i=0;i<n;i++) {cin>>a[i];}vector<int> vis(n+1,0);int ans=0;for(int i=0;i<min(31,n);i++){int maxOr=0,idx=-1;for(int j=0;j<n;j++){if(vis[j]) continue;if((a[j]|ans)>maxOr){maxOr=a[j]|ans;idx=j;}}vis[idx]=1;cout<<a[idx]<<" ";ans|=a[idx];}for(int i=0;i<n;i++){if(!vis[i]) cout<<a[i]<<" ";}cout<<endl;
}
signed main(){int t=1;cin>>t;while (t--) solve();return 0;
}

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Codeforces Round #827 (Div. 4) 题解记录

Codeforces Round #827 (Div. 4)

A

B

C

D

E

F

G

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