2022/1/1
2022/1/1
[ D. Keep the Average High]( Problem - D - Codeforces (Unofficial mirror site, accelerated for Chinese users) )
思路：
一、将a[i]的每个数-x，那么题目就转换成了，将a[i]分成若干个区间，每个区间内部的小区间之和都大于等于0，（跟求最大字段和有异曲同工之妙。
参考代码：
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int, int >
#define se second
#define pb push_back
#define pf push_front
#define si size()
#define db double
#define ls (p<<1)
#define rs (p<<1|1)
#define mid (t[p].l+t[p].r)/2
#define fi first
#define se second
using namespace std;
ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void Prin(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) Prin(x / 10);putchar(x % 10 + '0');}
const int qs=2e5+7;
ll T,n,a[qs];
int main(){
T=read();
while(T--){
n=read();
for(int i=1;i<=n;++i) a[i]=read();
ll x; x=read();
ll ans=0,y=1e9;
for(int i=1;i<=n;++i){
a[i]-=x;
int flag=0;
if(y+a[i]<0){
ans++; flag=1;
}
y=min(a[i],y+a[i]);
if(flag){
y=1e9;
}
}
cout<<n-ans<<"\n";
}
return 0;
}
二、dp
\(f[i][j]\)表示第i个数往前连续选j个数的区间长度
观察样例会发现，只有区间长度为0，1，2的情况为特殊情况，长度>=3的为一种情况
参考代码
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int, int >
#define se second
#define pb push_back
#define pf push_front
#define si size()
#define db double
#define ls (p<<1)
#define rs (p<<1|1)
#define mid (t[p].l+t[p].r)/2
#define fi first
#define se second
using namespace std;
ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void Prin(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) Prin(x / 10);putchar(x % 10 + '0');}
const int qs=2e5+7;
ll T,n,a[qs];
ll f[qs][4];
int main(){
T=read();
while(T--){
n=read();
for(int i=1;i<=n;++i) {
a[i]=read();
f[i][0]=f[i][1]=f[i][2]=f[i][3]=0;
}
ll x; x=read();
ll ans=0;
for(int i=1;i<=n;++i){
f[i][0]=max(max(f[i-1][0],f[i-1][1]),max(f[i-1][2],f[i-1][3]));
f[i][1]=f[i-1][0]+1;
if(i>1&&a[i]+a[i-1]>=2*x){
f[i][2]=f[i-1][1]+1;
if(i>2&&a[i]+a[i-1]+a[i-2]>=3*x&&a[i-1]+a[i-2]>=2*x){
f[i][3]=max(f[i-1][2]+1,f[i-1][3]+1);
}
}
}
cout<<max(max(f[n][0],f[n][1]),max(f[n][2],f[n][3]))<<"\n";
}
return 0;
}
[E. Lexicographically Small Enough]( Problem - E - Codeforces (Unofficial mirror site, accelerated for Chinese users) )
思路
首先把26个字母分类统计，考虑交换字母时，分两种情况，设当前字母是x：
- 找比x小的字母。枚举<=x的字母，找到一个下标最小且没被使用过的，直接前缀和记录答案。
- 等于x。在保存x的队列中取出队头，设当前下标为i，队头下标为j，那么总次数需要加上 j-1 的前缀和（即前j-1个没有被使用过的），再将j下标的值-1（标记为用过），并抛出队列
参考代码
#include<bits/stdc++.h>
#define ll long long
#define pii pair<char, int >
#define se second
#define pb push_back
#define pf push_front
#define si size()
#define db double
#define ls (p<<1)
#define rs (p<<1|1)
#define mid (t[p].l+t[p].r)/2
#define fi first
#define se second
using namespace std;
ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void Prin(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) Prin(x / 10);putchar(x % 10 + '0');}
const int qs=2e5+7;
int T,n;
string a,b;
queue<int> q[500];
int val[qs];
int lb(int x){
return (x&(-x));
}
void add(int x,int k){
for(;x<=n;x+=lb(x)) val[x]+=k;
}
ll ask(ll x){
ll ret=0;
for(;x>0;x-=lb(x)) ret+=val[x];
return ret;
}
int main(){
T=read();
while(T--){
n=read();
cin>>a; a=" "+a;
cin>>b; b=" "+b;
for(int i='a';i<='z';++i) {
while(q[i].si) q[i].pop();
}
for(int i=1;i<=n;++i) val[i]=0;
for(int i=1;i<=n;++i){
q[a[i]].push(i);
add(i,1);
}
ll ans=1e12,sum=0;
for(int i=1;i<=n;++i){
for(int j='a';j<b[i];++j){
if(q[j].si){
ll p=ask(q[j].front()-1);
ans=min(ans,sum+p);
}
}
if(!q[b[i]].si) break;
ll p=ask(q[b[i]].front()-1);
sum+=(p);
add(q[b[i]].front(),-1);
q[b[i]].pop();
}
if(ans==1e12) ans=-1;
cout<<ans<<"\n";
}
return 0;
}
[ C. Set or Decrease]( Problem - C - Codeforces (Unofficial mirror site, accelerated for Chinese users) )
思路
二分答案。
先排个序，接着二分次数，枚举变数组后面几个，其他次数全给a[1]。
参考代码
#include<bits/stdc++.h>
#define ll long long
#define pii pair<char, int >
#define se second
#define pb push_back
#define pf push_front
#define si size()
#define db double
#define ls (p<<1)
#define rs (p<<1|1)
#define fi first
#define se second
using namespace std;
ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void Prin(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) Prin(x / 10);putchar(x % 10 + '0');}
const int qs=2e5+7;
ll T,n,k,a[qs],sum;
int ck(ll x){
ll all=sum;
if(all-x<=k) return 1;
all-=a[1];
ll f=a[1];
for(int i=n;i>=2&&x>=(n-i+1);--i){
all-=a[i];
ll fp=all+(f-(x-(n-i+1)))*(n-i+2);
if(fp<=k) return 1;
}
return 0;
}
int main(){
T=read();
while(T--){
n=read(),k=read();
sum=0;
for(int i=1;i<=n;++i) a[i]=read(),sum+=a[i];
sort(a+1,a+1+n);
int l=0,r=1e12,ans;
while(l<=r){
int md=(l+r)/2;
if(ck(md)) r=md-1,ans=md;
else l=md+1;
}
cout<<ans<<"\n";
}
return 0;
}
