题目链接
官方视频讲解
A. Archmage
签到题
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define pi pair<int, int>
#define mk make_pair
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
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=1e3+10;
double x[N],y[N],z[N];
int n;
double dis(double x,double y,double z)
{
return sqrt(x * x + y * y + z * z);
}
int main()
{
int _ = read();while(_--)
{
ll n = read(), m = read(), x = read(), y = read();
if(y >= x){
if(n>=x) printf("%lld\n",m);
else printf("%lld\n",m-1);
continue;
}
ll sum = n + (m - 1) * y;
ll ans = sum / x;
ans = min(ans,m);
printf("%lld\n",ans);
}
}
B. Bamboo Leaf Rhapsody
签到题
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define pi pair<int, int>
#define mk make_pair
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
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=1e3+10;
double x[N],y[N],z[N];
int n;
double dis(double x,double y,double z)
{
return sqrt(x*x+y*y+z*z);
}
int main()
{
n=read();
rep(i,1,n) cin>>x[i]>>y[i]>>z[i];
double ans=dis(x[1],y[1],z[1]);
rep(i,1,n)
{
ans=min(ans,dis(x[i],y[i],z[i]));
}
printf("%.3f\n",ans);
}
C. Cheat Sheet
签到题
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define pi pair<int, int>
#define mk make_pair
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
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=1e3+10;
int n,m;
int a[N];
string t[N];
int main()
{
n = read(),m = read();
rep(i,1,m) cin>>t[i];
sort(t+1,t+1+m);
m=unique(t+1,t+1+m)-t-1;
rep(i,1,m) a[i]=t[i].size();
sort(a+1,a+1+m);
int ans=0;
rep(i,1,m)
{
//printf("i:%d sz:%d\n",i,a[i]);
if(n>=a[i]) n-=a[i],ans++;
else break;
if(i!=m) n--;
}
printf("%d\n",ans);
}
D. Disaster Recovery
这题我傻了，我先比较min(x,y)再比较max(x,y)了，粗略证明：
ai+bi 跟 aj+bj 比大小，先比大的 max(ai,bi)>max(aj,bj) 成立 又ai=ai-1+ai-2 那么aj+bj 最大最大也就只能得到ai 但我的ai还加了个bi。
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define rep(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;
struct node
{
int u,v;
pair<int,int>w;
bool operator <(const node &o)const{
return w<o.w;
}
}a[N];
int fa[N],du[N],n,m;
int fin(int x)
{
if(fa[x]!=x) fa[x]=fin(fa[x]);
return fa[x];
}
int main()
{
n=read(),m=read();
rep(i,1,n) fa[i]=i;
rep(i,1,m){
int u=read(),v=read();
a[i]={u,v,make_pair(max(u,v),min(u,v))};
}
sort(a+1,a+1+m);
rep(i,1,m)
{
int f1=fin(a[i].u);
int f2=fin(a[i].v);
if(f1==f2) continue;
du[a[i].u]++;
du[a[i].v]++;
fa[f1]=f2;
}
int ans=0;
rep(i,1,n) ans=max(ans,du[i]);
printf("%d\n",ans);
}
G. Gentle Jena
这题看b站视频补的。
1、单独考虑b[i]的贡献就是 bi*li*ri li 是 以bi为最小值时的左端点到i的距离长度
2、那么新增一个b[i] 对前面所有的小于当前b[i]的b[j]们 的贡献就是增加了一个 bj*lj
3、bj*lj*rj=>bj*lj*(rj+1) //这里 代码中用now维护
4、i新增的贡献就是li*bi
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define fi first
#define se second
#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 ll mod=998244353;
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=1e7+10;
ll n,p,x,y,z,b;
int main()
{
n=read(),p=read(),x=read(),y=read(),z=read(),b=read();
ll now=0,ans=0,sum=0;
// 单独考虑b[i]的贡献就是 bi*li*ri li 是 以bi为最小值时的左端点到i的距离长度
//那么新增一个b[i] 对前面所有的小于当前b[i]的b[j]们 的贡献就是增加了一个 bj*lj
//bj*lj*rj=>bj*lj*(rj+1) //这里 代码中用now维护
//i新增的贡献就是li*bi
for(int i=1;i<=n;++i){
ll len=1;
while(sta.size()&&sta.top().first>=b){
len+=sta.top().second;
now=(now-sta.top().first*sta.top().second%mod+mod)%mod;
sta.pop();
}
//printf("len:%lld now:%lld sum:%lld\n",len,now,sum);
sta.push(make_pair(b,len));
now=(now+b*len%mod)%mod;
sum=(sum+now)%mod;
//printf("len:%lld now:%lld sum:%lld\n\n",len,now,sum);
ans=ans^sum;
b=(x*sum%p+y*b%p+z)%p;
}
printf("%lld\n",ans);
}
H. Hay Mower
做法：时间复杂度可以达到：k*max(n,m) 你就知道怎么做了，维护每行 每列的最大时间就可以了。
#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;
const ll mod=998244353;
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;
ll a[N][N],n,m,k;
ll row[N],col[N];
void add(ll &x,ll y)
{
x=(x+y)%mod;
}
int main()
{
n=read(),m=read(),k=read();
rep(i,1,n) rep(j,1,m) a[i][j]=read()%mod;
ll ans=0;
while(k--)
{
char s[2];
scanf("%s",s);ll x=read(),t=read();
if(s[0]=='r'){
ll bs=row[x];
for(int i=1;i<=m;++i){
add(ans,(t-max(bs,col[i]))%mod*a[x][i]%mod);
}
row[x]=t;
}
else{
ll bs=col[x];
for(int i=1;i<=n;++i){
add(ans,(t-max(bs,row[i]))%mod*a[i][x]%mod);
}
col[x]=t;
}
//printf("%lld\n",ans);
}
printf("%lld\n",ans);
}
L. Lottery Tickets
做法：枚举后两个数，后两个数尽量的小，也就是大的数 尽量在高位输出就可以了。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define fi first
#define se second
#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 a[12],sum;
int main()
{
int _=read();while(_--)
{
sum=0;
rep(i,0,9) {
a[i]=read();sum+=a[i];
}
if(sum==a[0]){puts("0");continue;}
pair<int,int>ans={-1,-1};
rep(i,0,9){
rep(j,0,9){
if((i*10+j)%4!=0) continue;
if(!a[i]||!a[j]) continue;
if(i==j&&a[i]<2) continue;
///printf("i:%d j:%d\n",i,j);
if(ans.fi==-1){
ans={i,j};
continue;
}
if(max(i,j)<max(ans.fi,ans.se)){
ans={i,j};
}
else if(min(i,j)<min(ans.fi,ans.se)){
ans={i,j};
}
else if(max(i,j)==max(ans.fi,ans.se)&&min(i,j)==min(ans.fi,ans.se)){
///printf("i:%d j:%d fi:%d se:%d\n",i,j,ans.fi,ans.se);
ans={max(i,j),min(i,j)};
}
}
}
if(ans.fi==-1){
if(a[8]) puts("8");
else if(a[4]) puts("4");
else if(a[0]) puts("0");
else puts("-1");
}
else{
a[ans.fi]--;
a[ans.se]--;
for(int i=9;i>=0;--i){
while(a[i]--) printf("%d",i);
}
printf("%d%d\n",ans.fi,ans.se);
}
}
}