比赛链接

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A R

题目链接
题意：

题解：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5;
int a[N], p[N];
signed main() {int n, k;cin>>n>>k;string s;cin>>s;s=s+"P";int cnt=0;a[0]=-1;for (int i=0; i<=n; i++) {if(s[i]=='P') a[++cnt]=i;} int ans=0; for(int j=1; j<=cnt; j++) { int cntt=0;p[0]=0;for (int i=a[j-1]+1; i<a[j]; i++) {if (s[i]=='R') p[++cntt]=i+1-a[j-1]-1;if(cntt>=k) ans+=p[cntt-k+1];} }cout<<ans<<endl;return 0;
}

B 进制

题目链接
题意：

题解：

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=1e5+5;
const int M=1e9+7;
int a[N];
int b[13][N];
int n, q;
int p[13][N];
int maxx[4*N];
int sum[13][4*N];
ll ksm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%M;}a=a*a%M;p>>=1;}return res;}
void pushup(int id) {maxx[id] = max(maxx[id << 1], maxx[id << 1 | 1]);
}
void build(int id, int l, int r) {if (l == r) {maxx[id] = a[l];return;}int mid = (l + r) >> 1;build(id << 1, l, mid);build(id << 1 | 1, mid + 1, r);pushup(id);
}
void update(int id, int l, int r, int x, int v) {if (l == r) {maxx[id] = v;return;}int mid = (l + r) >> 1;if (x <= mid) {update(id << 1, l, mid, x, v);} else {update(id << 1 | 1, mid + 1, r, x, v);}pushup(id);
}
int query(int id, int l, int r, int x, int y) {if(x<=l&&r<=y) {return maxx[id];}int mid=(l+r)>>1;int ans=0;if(x<=mid) {ans=max(ans, query(id<<1, l, mid, x, y));}if(y>mid) {ans=max(ans, query(id<<1|1, mid+1, r, x, y));}return ans;
}
void pushup1(int id, int jin) {sum[jin][id] = (sum[jin][id << 1]+sum[jin][id << 1 | 1])%M;
}
void build1(int id, int l, int r, int jin) {if (l == r) {sum[jin][id] = b[jin][l]%M;return;}int mid = (l + r) >> 1;build1(id << 1, l, mid, jin);build1(id << 1 | 1, mid + 1, r, jin);pushup1(id, jin);
}
void update1(int id, int l, int r, int x, int v, int jin) {if (l == r) {sum[jin][id] = v%M;return;}int mid = (l + r) >> 1;if (x <= mid) {update1(id << 1, l, mid, x, v,jin);} else {update1(id << 1 | 1, mid + 1, r, x, v,jin);}pushup1(id, jin);
}
int query1(int id, int l, int r, int x, int y,int jin) {if(x<=l&&r<=y) {return sum[jin][id]%M;}int mid=(l+r)>>1;int ans=0;if(x<=mid) {ans=ans%M+query1(id<<1, l, mid, x, y,jin)%M;}if(y>mid) {ans=ans%M+query1(id<<1|1, mid+1, r, x, y,jin)%M;}ans%=M;return ans;
}
signed main() {cin>>n>>q;string s;cin>>s;s='.'+s;for(int i=n; i>=1; i--) {a[i]=(s[i]-'0');}for(int i=n; i>=1; i--) {for(int j=2; j<=10; j++) {b[j][i]=a[i]*ksm(j, n-i)%M;//cout<<i<<" "<<j<<" "<<b[j][i]<<endl;}}build(1, 1, n);for(int i=2; i<=10; i++) build1(1, 1, n, i);//cout<<"$$$"<<query1(1,1,n,1 ,3, 10)<<endl;while(q--) {int op, x, y;cin>>op>>x>>y;if(op==1) {update(1, 1, n, x, y);for(int i=2; i<=10; i++) update1(1, 1, n, x, y*ksm(i, n-x), i);}else {int dang=query(1, 1, n, x, y);if(dang==0) cout<<0<<endl;else {dang++;cout<<query1(1,1,n,x, y,dang)%M*ksm(ksm(dang, n-y), M-2)%M<<endl;}}}return 0;
}

C 蓝彗星

题目链接
题意：
题解：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5;
int a[N], b[N], c[N];
signed main() {int n, t;cin>>n>>t;string s;cin>>s;s='.'+s;for(int i=1; i<=n; i++) {int x; cin>>x;if(s[i]=='B') b[x]++, b[x+t]--;else c[x]++, c[x+t]--;}for(int i=1; i<=N; i++) {b[i]+=b[i-1];c[i]+=c[i-1];//cout<<i<<" "<<b[i]<<" "<<c[i]<<endl;}int ans=0;for(int i=1; i<=N; i++) {if(b[i]&&!c[i]) ans++;}cout<<ans<<endl;return 0;
}

D 雪色光晕

题目链接
题意：

题解：

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=2e5+5;
const int M=1e9+7;
double dis(double xx, double yy, double x1, double y2) {return sqrt((xx-x1)*(xx-x1)+(yy-y2)*(yy-y2));
}
double Pd(double x, double y, double x1, double y1, double x2, double y2){double cross = (x2 - x1) * (x - x1) + (y2 - y1) * (y - y1);if (cross <= 0) return sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));double d2 = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);if (cross >= d2) return sqrt((x - x2) * (x - x2) + (y - y2) * (y - y2));double r = cross / d2;double px = x1 + (x2 - x1) * r;double py = y1 + (y2 - y1) * r;return sqrt((x - px) * (x - px) + (py - y) * (py - y));
}
signed main() {int n;cin>>n;double x0, y0, X, Y;cin>>x0>>y0>>X>>Y;double minn=dis(x0,y0,X,Y);for(int i=1; i<=n; i++) {double x, y;cin>>x>>y;minn=min(minn, Pd(X, Y, x0,y0, x0+x, y0+y));x0+=x;y0+=y;}cout<<fixed<<setprecision(8)<<minn<<endl;return 0;
}

E 真假签到题

题目链接
题意：

题解：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5; 
signed main() {int x;cin>>x;cout<<x<<endl;return 0;
}

F 小红的记谱法

题目链接
题意：

题解：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5;
int a[N], b[N], c[N];
map<char, int> mp;
signed main() {string pp="CDEFGAB";for(int i=0; i<pp.size(); i++) mp[pp[i]]=i+1;string s;cin>>s;int n=s.size();int p=0;for(int i=0; i<n; i++) {// cout<<i<<" "<<p<<endl;if(s[i]=='<') p++;else if(s[i]=='>') p--;else {//cout<<i<<" "<<p<<endl;cout<<mp[s[i]];if(p==0) continue;else if(p>0) {for(int j=1; j<=p; j++) cout<<'.';}else for(int j=1; j<=-p; j++) cout<<'*';}}cout<<endl;return 0;
}

G 子序列权值乘积

题目链接

题意：

题解：

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=2e5+5;
const int M=1e9+7;
const int MM=1e9+6;
int a[N], b[N];
int dp[1005][1005];
ll ksm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%M;}a=a*a%M;p>>=1;}return res%M;}
ll ksmm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%MM;}a=a*a%MM;p>>=1;}return res%MM;}
signed main() {int n;cin>>n;for(int i=1; i<=n; i++) cin>>a[i];sort(a+1, a+1+n); int ans=1;for(int i=1; i<=n; i++) {int b=ksmm(2, i-1)+ksmm(2, n-i);ans=ans%M*ksm(a[i], b)%M;ans%=M;}cout<<ans%M<<endl;return 0;
}

H 真真真真真签到题

题目链接

题意：

题解：

#include<bits/stdc++.h>
#define ihnt long long
using namespace std;
const int N=2e5+5; 
signed main() {double x;cin>>x;cout<<fixed<<setprecision(7)<<x/sqrt(3)*2*x/sqrt(3)*2*x/sqrt(3)*2<<endl;return 0;
}

I 爆炸的符卡洋洋洒洒

题目链接

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=2e5+5;
const int M=1e9+7;
int a[N], b[N];
int dp[1005][1005];
signed main() {int n, k;cin>>n>>k;for(int i=1; i<=n; i++) {cin>>a[i]>>b[i];a[i]%=k;}for(int i=1; i<=n; i++) {for(int j=k-1; j>=0; j--) {int p=(j-a[i]+k)%k;dp[j][i]=max(dp[j][i-1], dp[j][i]); if(dp[p][i-1]||p==0)dp[j][i]=max(dp[p][i-1]+b[i], dp[j][i]);  }}int maxx=dp[0][n];if(maxx<=0) cout<<-1<<endl;else cout<<maxx<<endl;return 0;
}

J 区间合数的最小公倍数

题目链接

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=2e5+5;
const int M=1e9+7;
int p[N];
bool check[N],vis[N];
int n, m;
void init() {int t;check[1]=true;for(int i=2;i<=n;i++){if(!check[i])p[++p[0]]=i;for(int j=1;j<=p[0];j++){t=i*p[j];if(t>n)break;check[t]=true;if(i%p[j]==0)break;}}
}
ll ksm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%M;}a=a*a%M;p>>=1;}return res;}
int mp[N];
signed main() {n=30001;init();int l, r;cin>>l>>r;bool q=0; for(int i=l; i<=r; i++) {int num=i;int m=0;for(int j=1; j<=p[0]; j++) {while(num%p[j]==0) {num/=p[j];m++;}}if(m>=2) q=1;else continue; num=i;for(int j=1; j<=p[0]; j++) {int sum=0;while(num%p[j]==0) {num/=p[j];sum++;}mp[p[j]]=max(sum, mp[p[j]]);}}if(!q) {cout<<-1<<endl;return 0;}int ans=1;for(int i=1; i<=p[0]; i++) { ans=ans*ksm(p[i], mp[p[i]])%M;ans%=M;}cout<<ans<<endl;return 0;
}

K 小红的真真假假签到题题

题目链接

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5;
signed main() {int x;cin>>x;string s;while(x) {s=char('0'+x%2)+s;x/=2;}s=s+s;int ans=0;int p=1;for(int i=0; i<s.size(); i++) {ans=ans*2+(s[i]-'0');}cout<<ans<<endl;return 0;
}

L 在这冷漠的世界里光光哭哭【待补】

题目链接

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总结

Qwq