题目链接
A-俄罗斯方块
做法：简单模拟
#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 n;
int a[15][15];
int row[15];
int main()
{
n=read();
rep(i,1,10) row[i]=10;
rep(j,1,n)
{
int ty=read(),x=read();
if(ty==1){
//(i-1,x) (i-1,x+1)
//(i,x) (i,x+1)
per(i,row[x],1){
if(!a[i][x]&&!a[i][x+1]&&!a[i-1][x]&&!a[i-1][x+1]){
a[i][x]=a[i][x+1]=a[i-1][x]=a[i-1][x+1]=1;
row[x]=i-2;
row[x+1]=i-2;
break;
}
}
}
else if(ty==2){
//a[i-1][x]
//a[i][x] a[i][x+1] a[i][x+2]
per(i,row[x],1)
if(!a[i][x]&&!a[i][x+1]&&!a[i][x+2]&&!a[i-1][x]){
a[i][x]=a[i][x+1]=a[i][x+2]=a[i-1][x]=1;
row[x]=i-2;
row[x+1]=i-1;
row[x+2]=i-1;
break;
}
}
else if(ty==3){
//(i,x) (i,x+1) (i,x+2) (i,x+3)
per(i,row[x],1)
if(!a[i][x]&&!a[i][x+1]&&!a[i][x+2]&&!a[i][x+3]){
a[i][x]=a[i][x+1]=a[i][x+2]=a[i][x+3]=1;
row[x]=i-1;
row[x+1]=i-1;
row[x+2]=i-1;
row[x+3]=i-1;
break;
}
}
else if(ty==4){
// (i-1,x+1)
//(i,x)(i,x+1)(i,x+2)
per(i,row[x],1)
if(!a[i][x]&&!a[i][x+1]&&!a[i][x+2]&&!a[i-1][x+1]){
//printf("i:%d x:%d\n",i,x);
a[i][x]=a[i][x+1]=a[i][x+2]=a[i-1][x+1]=1;
row[x]=i-1;
row[x+1]=i-2;
row[x+2]=i-1;
break;
}
}
}
//printf("dddd%d\n",a[10][1]);
rep(i,1,10)
{
rep(j,1,10) printf("%d%c",a[i][j],j==10?'\n':' ');
//puts("");
}
}
/*
2 1
2 2
1 1
3 3
*/
C-港口
题目言外之意就是使得数组的差分数组为0参考做法来自：牛客
#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=1e5+10;
ll a[N],n;
int main()
{
n=read();
rep(i,1,n) a[i]=read();
ll ma1 = 0,ma2 = 0;
for(int i=1;i<=n;++i)
{
if(i == 1) continue;
ll tmp = a[i]-a[i-1];
if(tmp > 0) ma1 += tmp;
else ma2 -= tmp;//减负数其实就是加
//printf("ma1:%lld ma2:%lld tmp:%lld\n",ma1,ma2,tmp);
}
ll ans = max(ma1,ma2);
printf("%lld\n",ans);
return 0;
}
/*
7
0 2 2 2 3 3 1 1
ans:2
*/
D-构造数组
做法：离线倒过来模拟，主席树维护一下空的位置即可，代码下方带一个不错的数据
#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=5e5+10;
int in[N],a[N],n,ans[N],sum[4*N];
int qu(int id,int l,int r,int k)
{
if(l==r) return l;
int mid=l+r>>1;
//printf("id:%d su:%d k:%d\n",id,sum[id<<1],k);
if(k<=sum[id<<1]) return qu(id<<1,l,mid,k);
return qu(id<<1|1,mid+1,r,k-sum[id<<1]);
}
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);
}
void bu(int id,int l,int r)
{
if(l==r) {
sum[id]++;
return ;
}
int mid=l+r>>1;
bu(id<<1,l,mid);
bu(id<<1|1,mid+1,r);
sum[id]=sum[id<<1]+sum[id<<1|1];
}
int main()
{
cin>>n;
rep(i,1,n) cin>>in[i]>>a[i];
bu(1,1,n);
per(i,n,1)
{
int id=qu(1,1,n,in[i]);
//printf("id:%d\n",id);
ans[id]=a[i];
up(1,1,n,id);
}
rep(i,1,n) printf("%d ",ans[i]);
}
/*
7
1 1
2 2
3 3
2 4
3 5
4 6
5 7
ans:
1 4 5 6 7 2 3
*/
E-简单的线性代数
知识点来自：百度
#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=1e3+10;
int a[N][N];
int n;
ll k;
int main()
{
n=read(),k=read();
rep(i,1,n) rep(j,1,n) a[i][j]=read();
rep(i,1,n) rep(j,1,n)
{
if(i==j) a[i][j]=1-a[i][j];
else a[i][j]=-a[i][j];
}
rep(i,1,n){
rep(j,1,n) printf("%d ",a[i][j]);
puts("");
}
}
F-集合操作
做法：离线离散化每个x 和 x+1 添加和删除线段树处理就可以了
然后 查询就是 x[i] 到n的最小值即可，维护区间最小值。
#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;
int mi[8*N],len,X[2*N],n,id[N],x[N];
int get(int x)
{
return lower_bound(X+1,X+1+len,x)-X;
}
void bu(int id,int l,int r)
{
if(l==r) {
mi[id]=X[l];
return ;
}
int mid=l+r>>1;
bu(id<<1,l,mid);
bu(id<<1|1,mid+1,r);
mi[id]=min(mi[id<<1],mi[id<<1|1]);
//printf("id:%d mi:%d id<<1:%d id<<1|1:%d\n",id,mi[id],mi[id<<1],mi[id<<1|1]);
}
void up(int id,int l,int r,int pos,int val)
{
if(l==r) {
mi[id]=val;
return ;
}
int mid=l+r>>1;
if(pos<=mid) up(id<<1,l,mid,pos,val);
else up(id<<1|1,mid+1,r,pos,val);
mi[id]=min(mi[id<<1],mi[id<<1|1]);
// printf("id:%d mi:%d id<<1:%d id<<1|1:%d\n",id,mi[id],mi[id<<1],mi[id<<1|1]);
}
int qu(int id,int l,int r,int ql,int qr)
{
if(ql<=l&&r<=qr) return mi[id];
int mid=l+r>>1;
int ans=1e9+10;
if(ql<=mid) ans=qu(id<<1,l,mid,ql,qr);
if(qr>mid) ans=min(ans, qu(id<<1|1,mid+1,r,ql,qr));
return ans;
}
int main()
{
n=read();
rep(i,1,n)
{
id[i]=read();
x[i]=read();
X[++len]=x[i];
X[++len]=x[i]+1;
}
sort(X+1,X+1+len);
len=unique(X+1,X+1+len)-X-1;
//rep(i,1,n) printf("%d ",X[i]);
//puts("");
//printf("len:%d\n",len);
bu(1,1,len);
set<int>st;
rep(i,1,n)
{
//printf("i:%d mi:%d\n",i,mi[1]);
x[i]=get(x[i]);
//printf("x:%d\n",x[i]);
if(id[i]==1){
if(st.find(x[i])!=st.end()) continue;
up(1,1,len,x[i],1e9+10);
st.insert(x[i]);
}
if(id[i]==2){
if(st.find(x[i])==st.end()) continue;
//printf("插入x:%d\n",X[x[i]]);
up(1,1,len,x[i],X[x[i]]);
st.erase(x[i]);
}
if(id[i]==3){
//printf("l:%d r:%d\n",x[i],len);
int ans=qu(1,1,len,x[i],len);
printf("%d\n",ans);
}
}
}G-数位操作1
做法：刚开始想复杂了，唯一分解去了，其实就是枚举从9到2能否整除即可，注意个位数的n最小应该是1n
#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=1e5+10;
int ans[N],len;
int main()
{
ll n;
while(~scanf("%lld",&n)){
if(n<=9){
printf("1%d\n",n);
continue;
}
len=0;
int flag=1;
for(int i=9;i>=2;--i){
if(n%i==0){
while(n%i==0) ans[++len]=i,n=n/i;
}
}
if(n!=1){
if(n>=10) flag=0;
else ans[++len]=n;
}
if(flag==0){puts("-1");continue;}
//printf("len:%d\n",len);
sort(ans+1,ans+1+len);
rep(i,1,len) printf("%d",ans[i]);
puts("");
}
}
H-数位操作2
类原题：去年三月份某一场牛客练习赛，知道为啥记得这么清楚，那是我做的第一个dp难题。
做法：n只有50
设dp[i][s][j][k] 为前i位 各位数和为s 后两位是 j k 的方案数，简单推一推就可以了。
#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;
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=1000009;
const int N=55;
int dp[55][600][10][10],n,s,x;
void add(int &x,int y)
{
x=(x+y)%mod;
}
int main()
{
scanf("%d%d%d",&n,&s,&x);
if(s>=500){
printf("0");
return 0;
}
for(int i=0;i<=9;++i)
for(int j=0;j<=9;++j)
dp[2][i+j][i][j]=1;
for(int i=3;i<=n;++i){
rep(ss,0,500)
rep(j,0,9)
rep(k,0,9)
rep(c,0,9)
{
if((j*100+k*10+c)%x==0){
//printf("j:%d\n",j);
add(dp[i][ss+c][k][c],dp[i-1][ss][j][k]);
}
}
}
int ans=0;
rep(i,0,9)
rep(j,0,9) add(ans,dp[n][s][i][j]);
printf("%d\n",ans);
}
I-配对
另写了博客：最下面：博客
K-签个到
做法：线段树维护最大值最小值，其实做复杂了，完全维护前后缀的最大值最小值就可以了。
#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;
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=1e5+10;
ll mx[4*N],a[N],mi[4*N];
int n;
ll m;
void build(int id,int l,int r)
{
if(l==r){a[l]=read();mi[id]=mx[id]=a[l];return ;}
int mid=l+r>>1;
build(id<<1,l,mid);
build(id<<1|1,mid+1,r);
mx[id]=max(mx[id<<1],mx[id<<1|1]);
mi[id]=min(mi[id<<1],mi[id<<1|1]);
}
void up(int id,int l,int r,int pos,ll val)
{
if(l==r){
mi[id]=mx[id]=val;
return ;
}
int mid=l+r>>1;
if(pos<=mid) up(id<<1,l,mid,pos,val);
else up(id<<1|1,mid+1,r,pos,val);
mx[id]=max(mx[id<<1],mx[id<<1|1]);
mi[id]=min(mi[id<<1],mi[id<<1|1]);
}
int main()
{
n=read(),m=read();
build(1,1,n);
ll ans=0;
rep(i,1,n)
{
ll t=a[i];
up(1,1,n,i,t+m*i);
ans=max(ans,mx[1]-mi[1]);
up(1,1,n,i,t-m*i);
ans=max(ans,mx[1]-mi[1]);
up(1,1,n,i,t);
}
printf("%lld\n",ans);
}
L-20200524
假设倍数p=20200524
这类题型有O(plog(p))的公式做法：博客1
也有O(p^2)的同余做法：博客2
但是在这 上面的方法均超时
唯独不会O(p)容斥的做法，今天受教了~
献上代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=20200524;
vector<int> factor;
vector<int> f;
int main(){
for(int i=1;i*i<=mod;i++){
if(mod%i==0){
factor.push_back(i);
if(i*i!=mod) factor.push_back(mod/i);
}
}
sort(factor.begin(),factor.end());
int cnt=factor.size();
f.resize(cnt);
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=0;i<cnt;i++) f[i]=n/factor[i];//包含factor[i]的个数
for(int i=cnt-2;i>=0;i--){
for(int j=i+1;j<cnt;j++){
if(factor[j]%factor[i]==0){
f[i]-=f[j];//容斥一下
}
}
}
ll ans=0;
for(int i=0;i<cnt;i++){
ans+=1LL*f[i]*(m/(mod/factor[i]));//统计答案 最小*facator[i]是mod的整数倍
}
cout<<ans<<endl;
}
return 0;
}持续待补......