[ABC185] AtCoder Beginner Contest 185
| Task Name | Time Limit | Memory Limit | ||
|---|---|---|---|---|
| A | ABC Preparation | 2 sec | 1024 MB | Submit |
| B | Smartphone Addiction | 2 sec | 1024 MB | Submit |
| C | Duodecim Ferra | 2 sec | 1024 MB | Submit |
| D | Stamp | 2 sec | 1024 MB | Submit |
| E | Sequence Matching | 2 sec | 1024 MB | Submit |
| F | Range Xor Query | 3 sec | 1024 MB | Submit |
A
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int INF=0x3f3f3f3f;
int main()
{
// freopen("1.in","r",stdin);
int a,b,c,d;
cin>>a>>b>>c>>d;
a=min(a,b);
a=min(a,c);
a=min(a,d);
cout<<a<<endl;
return 0;
}
B
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=1024;
int n,m,t;
int main()
{
// freopen("1.in","r",stdin);
int i;
int x,y;
int last=0;
cin>>n>>m>>t;
int lit=n;
for(i=1;i<=m;i++) {
cin>>x>>y;
n-=(x-last);
if(n<=0) {
printf("No\n");
return 0;
}
n+=(y-x);
if(n>lit) n=lit;
last=y;
}
n-=(t-last);
if(n<=0) printf("No\n");
else printf("Yes\n");
return 0;
}
C
\(C_{L-1}^{11}\)
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
// C(L-1,11)
const int N=256;
LL c[N][N];
int main()
{
// freopen("1.in","r",stdin);
int i,j;
for(i=0;i<=N-1;i++)
for(j=0;j<=i;j++) {
if(j==0||j==i) c[i][j]=1;
else c[i][j]=c[i-1][j]+c[i-1][j-1];
}
int n;
cin>>n;
cout<<c[n-1][11];
return 0;
}
D
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=4e5+5,INF=2e9;
int n,m,k=INF;
int a[N];
int paint(int len)
{
return len/k+(len%k>0);
}
int main()
{
/// freopen("1.in","r",stdin);
int i;
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)
scanf("%d",&a[i]);
sort(a+1,a+m+1);
a[++m]=n+1;
for(i=1;i<=m;i++) {
if(a[i]==a[i-1]+1) continue;
k=min(k,a[i]-a[i-1]-1);
}
int ans=0;
for(i=1;i<=m;i++) {
if(a[i]==a[i-1]+1) continue;
else ans+=paint(a[i]-a[i-1]-1);
}
cout<<ans<<endl;
return 0;
}
E
\(n^3\) 竟然可过，这 Atcoder 神机。
不过后来又被我优化成了 \(O(n^2)\) 了。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=1024,INF=0x3f3f3f3f;
int f[N][N];
int n,m;
int a[N],b[N];
int g1[N],g2[N];
int main()
{
// freopen("1.in","r",stdin);
int i,j;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++) scanf("%d",&a[i]);
for(i=1;i<=m;i++) scanf("%d",&b[i]);
memset(f,0x3f,sizeof f);
int ans=INF;
f[0][0]=0;
for(j=0;j<=m;j++) g2[j]=f[0][j]-0-1;
for(i=1;i<=n;i++) {
g1[0]=f[i-1][0]-0-1;
for(j=1;j<=m;j++)
g1[j]=min(g1[j-1],f[i-1][j]-j-1);
for(j=1;j<=m;j++) {
// f[i][j]= min{ f[k][p]+(j-p-1)+(i-k-1)} + (a[i]!=a[j])
// f[i-1][j-1] = max { f[k][p]+(j-p-1)+(i-k-1) ) - 2 + (a[i-1]!=a[j-1]);
f[i][j]=f[i-1][j-1]-(a[i-1]!=b[j-1])+2+(a[i]!=b[j]);
// else f[i-1][k from 0 to j-1]
// for(int k=0;k<=j-1;k++)
// f[i][j]=min(f[i][j],f[i-1][k]+(j-k-1)+(a[i]!=b[j]));
f[i][j]=min(f[i][j],g1[j-1]+j+(a[i]!=b[j]));
// for(int k=0;k<=i-1;k++)
// f[i][j]=min(f[i][j],f[k][j-1]+i-k-1+(a[i]!=b[j]));
f[i][j]=min(f[i][j],g2[j-1]+i+(a[i]!=b[j]));
ans=min(ans,f[i][j]+(n-i)+(m-j));
}
for(j=0;j<=m;j++)
g2[j]=min(g2[j],f[i][j]-i-1);
}
cout<<ans<<endl;
return 0;
}
话说我状态定义想偏了，要是换一个可能更好些。
F
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=3e5+5;
int n,m;
struct Bits
{
int c[N];
inline int lowbit(int x) { return x&(-x); }
void Update(int x,int y)
{
for(;x<=n;x+=lowbit(x))
c[x]^=y;
}
int Sum(int x)
{
int res=0;
for(;x>=1;x-=lowbit(x))
res^=c[x];
return res;
}
}C;
int main()
{
// freopen("1.in","r",stdin);
int i;
int opt,x,y;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++) {
scanf("%d",&x);
C.Update(i,x);
}
while(m--) {
scanf("%d%d%d",&opt,&x,&y);
if(opt==1) C.Update(x,y);
else printf("%d\n",C.Sum(x-1)^C.Sum(y));
}
return 0;
}
维护一个前缀异或和。
