题目链接
A-牛牛爱学习
做法：二分天数mid，最大的mid分发给mid天，剩余的尽量插在mid天，判断是否能达到m即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e6+10;
ll sum[N],a[N],m;
int n;
bool cmp(ll x,ll y)
{
return x>y;
}
int run(int mid)
{
ll s=0;
for(int k=0;mid*k+1<=n;++k){
for(int j=1,i=mid*k+1;j<=mid;++j,++i){
if(a[i]-k>0) s+=a[i]-k;
}
}
//if(mid==1) printf(":%lld\n",s);
return s>=m;
}
int main()
{
n=read(),m=read();
rep(i,1,n) a[i]=read();
sort(a+1,a+1+n,cmp);
int l=1,r=n;
int ans=1e9;
while(l<=r)
{
int mid=l+r>>1;
if(run(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
if(ans==1e9) ans=-1;
printf("%d\n",ans);
}B-牛牛爱数学
做法：求根公式简单化简下，答案就是b*c/a 判断是否能整除即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
int main()
{
int _=read();while(_--)
{
ll a=read(),b=read(),c=read();
ll fz=b*c;
if(fz%a==0) printf("%lld\n",fz/a);
else puts("-1");
}
}C-牛牛种花
做法:偏序的经典问题。对一维排序，另一维 用线段树或者树状数组维护前缀和即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e5+10;
struct node
{
int x,y,ty,id;
}a[N];
int X[N],Y[N],lx,ly,len,ans[N],sum[4*N];
bool cmp(node a,node b)
{
if(a.x!=b.x) return a.x<b.x;
return a.y<b.y;
}
int qu(int id,int l,int r,int ql,int qr)
{
if(ql<=l&&r<=qr) return sum[id];
int mid=l+r>>1;
int ans=0;
if(ql<=mid) ans+=qu(id<<1,l,mid,ql,qr);
if(qr>mid) ans+=qu(id<<1|1,mid+1,r,ql,qr);
return ans;
}
void up(int id,int l,int r,int pos)
{
sum[id]++;
if(l==r) return ;
int mid=l+r>>1;
if(pos<=mid) up(id<<1,l,mid,pos);
else up(id<<1|1,mid+1,r,pos);
}
int main()
{
int n=read(),m=read();
rep(i,1,n)
{
int x=read(),y=read();
X[++lx]=x;
Y[++ly]=y;
a[++len]={x,y,1};
}
rep(i,1,m)
{
int x=read(),y=read();
X[++lx]=x;
Y[++ly]=y;
a[++len]={x,y,0,i};
}
sort(X+1,X+1+lx);
sort(Y+1,Y+1+ly);
lx=unique(X+1,X+1+lx)-X-1;
ly=unique(Y+1,Y+1+ly)-Y-1;
//printf("lx:%d ly:%d\n",lx,ly);
rep(i,1,len){
a[i].x=lower_bound(X+1,X+1+lx,a[i].x)-X;
a[i].y=lower_bound(Y+1,Y+1+ly,a[i].y)-Y;
}
//puts("***");
sort(a+1,a+1+len,cmp);
//rep(i,1,len)
rep(i,1,len)
{
//printf("i:%d\n",i);
if(a[i].ty==0){
ans[a[i].id]=qu(1,1,ly,1,a[i].y);
}
else{
up(1,1,ly,a[i].y);
}
}
rep(i,1,m) printf("%d\n",ans[i]);
}D-失忆药水
做法: 其实很简单的一道题，二分图是不含奇环的图，所以就是完全图的边数减去 最佳二分图的边数即可。
二分图边数怎么算？一边是n/2 另一边是n-n/2 两边点数相乘即可。
#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10,M=2e7+10,mod=998244353;
int n,m;
long long dp[N];
int main()
{
dp[2]=1;
for(int i=3;i<N;i++)dp[i]=dp[i-1]+(i)/2;
while(~scanf("%d",&n)){
printf("%lld\n",1ll*n*(n-1)/2-dp[n]);
}
return 0;
}E-牛牛走迷宫
做法：简单的输出路径的bfs
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=5e2+10,inf=0x3f3f3f3f;
char s[N][N];
int n,m,dir[4][2]={1,0,0,-1,0,1,-1,0};
int vs[N][N];
pair<int,int> ans[N][N];
void bfs()
{
memset(vs,inf,sizeof(vs));
queue<pair<int,int> >que;
que.push({1,1});
vs[1][1]=0;
while(que.size()){
auto now=que.front();que.pop();
for(int i=0;i<4;++i){
int x=now.first+dir[i][0];
int y=now.second+dir[i][1];
if(x<1||y<1||x>n||y>m||s[x][y]=='1') continue;
if(vs[x][y]<=vs[now.first][now.second]+1) continue;
//printf("x:%d y:%d\n",x,y);
ans[x][y]={now.first,now.second};
//printf("x:%d y:%d fi:%d se:%d\n",x,y,ans[x][y].first,ans[x][y].second);
vs[x][y]=vs[now.first][now.second]+1;
que.push({x,y});
}
}
}
int main()
{
n=read(),m=read();
rep(i,1,n) scanf("%s",s[i]+1);
bfs();
if(vs[n][m]==inf)
puts("-1");
else{
int x=n,y=m;
stack<char>sta;
while(x!=1||y!=1){
int nx=ans[x][y].first;
int ny=ans[x][y].second;
if(x==nx+1) sta.push('D');
if(y==ny+1) sta.push('R');
if(y==ny-1) sta.push('L');
if(x==nx-1) sta.push('U');
x=nx,y=ny;
}
printf("%d\n",sta.size());
while(sta.size()){
printf("%c",sta.top());sta.pop();
}
//sta.push(s[1][1]);
}
}F-牛牛的序列
做法：不太会，看的官方题解：
最后答案就是求区间的奇数平方数及其二次幂的个数。
[1,n] 的区间平方数个数
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf=1000000000;
const double pi=acos(-1);
ll num(ll x)
{
ll l=0,r=x,ans;
while(l<=r)
{
ll p=(l+r)/2;
if(p*p<=x)l=p+1,ans=p;
else r=p-1;
}
return l;
}
int main()
{
//for(int i=1;i<=1000;i++)printf("i=%d %d\n",i,sum(i));
ll T;
scanf("%lld",&T);
while(T--)
{
ll a,b;
scanf("%lld%lld",&a,&b);
if(a>b)swap(a,b);
ll ans=num(b)-num(a-1)+num(b/2)-num((a+1)/2-1);
//printf("ans=%lld\n",ans);
printf("%lld\n",ans%2);
}
return 0;
}
G-牛牛爱几何
做法：简单几何题
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const double pi=acos(-1);
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
int main()
{
double n;
while(cin>>n)
{
double r=n/2;
double ans=pi*r*r/4-r*r/2;
ans*=8;
printf("%.6f\n",ans);
}
}I-恶魔果实
做法：建有向图，然后求每个数值的能到达数的个数，组合数乘起来就是答案了。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int mod=1e4+7;
int x,n,vis[12],vs[12][12],dp[12];
vector<int>G[12];
int run(int u)
{
if(dp[u]!=0) return dp[u];
memset(vis,0,sizeof(vis));
int ans=0;
queue<int>que;que.push(u);
while(que.size()){
int now=que.front();que.pop();
ans++;vis[now]=1;
for(int v:G[now]) if(!vis[v]) {
que.push(v);
vis[v]=1;
}
}
dp[u]=ans;
return ans;
}
int main()
{
x=read(),n=read();
rep(i,1,n)
{
int u=read(),v=read();
if(vs[u][v]) continue;
vs[u][v]=1;
G[u].push_back(v);
}
int ans=1;
while(x)
{
int d=x%10;
ans=ans*run(d)%mod;
x=x/10;
}
printf("%d\n",ans);
}J-牛牛喜欢字符串
做法：水题，记录每个串内的位置字符个数，总个数减去最大的数即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const double pi=acos(-1);
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e6+10;
char s[N];
int dp[N][26],n,k;
int main()
{
n=read(),k=read();
scanf("%s",s+1);
int tot=n/k;
for(int i=1;i+k-1<=n;i=i+k){
for(int j=1,t=i;j<=k;++j,++t){
dp[j][s[t]-'a']++;
}
}
ll ans=0;
for(int i=1;i<=k;++i){
int mx=0;
for(int j=0;j<26;++j) mx=max(mx,dp[i][j]);
ans+=tot-mx;
}
printf("%lld\n",ans);
}