CFround#380 div2

题目链接:http://codeforces.com/contest/738

A题:SB题。

B题:SB题。

C题:二分。

D题:贪心。

E题:乱搞。

F题:设f[i][j][k]代表甲先手,左边消去了i个,右边消去了i+j个,上一次消去k个时答案。g[i][j][k]乙先手。可以证明j<=sqrt(2n),k<=sqrt(2n)。o(n^2)。

代码:

A:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 105
using namespace std;
int n;
char s[maxn];
bool check(int x){if(x>n-2)return 0;return s[x]=='o'&&s[x+1]=='g'&&s[x+2]=='o';}
int main(){
  scanf("%d",&n);scanf("%s",s+1);
  for(int i=1;i<=n;){
    if(check(i)){
      if(check(i+2))i+=2;
      else putchar('*'),putchar('*'),putchar('*'),i+=3;
    }else putchar(s[i]),i++;
  }
  puts("");
  return 0;
}

B题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1005
using namespace std;
int n,ans,m,a[maxn][maxn],sum[maxn][maxn];
int read(){
  int x=0,f=1;char ch;
  for(ch=getchar();ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
  for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
  return x*f;
}
int main(){
  n=read();m=read();
  for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)a[i][j]=read();
  for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j];
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)if(!a[i][j]){
	if(sum[i-1][j]-sum[i-1][j-1])ans++;
	if(sum[i][j-1]-sum[i-1][j-1])ans++;
	if(sum[n][j]-sum[n][j-1]-sum[i][j]+sum[i][j-1])ans++;
	if(sum[i][m]-sum[i][j]-sum[i-1][m]+sum[i-1][j])ans++;
    }
  printf("%d\n",ans);
  return 0;
}

C题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 200005
using namespace std;
int n,k,s,t,tot,a[maxn],c[maxn],v[maxn];
int read(){
  int x=0,f=1;char ch;
  for(ch=getchar();ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
  for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
  return x*f;
}
bool check(int x){
  int sum=0;
  for(int i=1;i<=tot;i++){
    if(x<a[i]-a[i-1])return 0;
    sum+=max(a[i]-a[i-1],2*a[i]-2*a[i-1]-(x-a[i]+a[i-1]));
  }
  if(sum>t)return 0;else return 1;
}
int main(){
  n=read();k=read();s=read();t=read();
  for(int i=1;i<=n;i++)c[i]=read(),v[i]=read();
  for(int i=1;i<=k;i++)a[++tot]=read();a[++tot]=s;
  sort(a+1,a+tot+1);
  int l=0,r=1e9,mid;
  while(l<=r){
    mid=(l+r)>>1;
    if(check(mid))r=mid-1;
    else l=mid+1;
  }
  int ans=2e9;
  for(int i=1;i<=n;i++)if(v[i]>r&&c[i]<ans)ans=c[i];
  if(ans!=2e9)printf("%d\n",ans);else puts("-1");
  return 0;
}

D题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 200005
using namespace std;
int n,a,b,k,tot,ans[maxn];
char s[maxn];
int main(){
  scanf("%d%d%d%d\n",&n,&a,&b,&k);scanf("%s",s+1);
  int pre=0;
  for(int i=1;i<=n;i++){
    if(s[i]=='1'){
      for(int j=pre+b;j<i;j+=b)ans[++tot]=j;
      pre=i;
    }
  }
  for(int j=pre+b;j<=n;j+=b)ans[++tot]=j;
  printf("%d\n",tot-a+1);
  for(int i=a;i<=tot;i++)printf("%d ",ans[i]);puts("");
  return 0;
}

E题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 200005
using namespace std;
int n,s,a[maxn],ans;
int read(){
  int x=0,f=1;char ch;
  for(ch=getchar();ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
  for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
  return x*f;
}
int main(){
  n=read();s=read();
  for(int i=1;i<=n;i++){
    a[i]=read();
    if(i==s&&a[i])a[i]=0,ans++;
    if(i!=s&&!a[i])a[i]=1e9;
  }
  sort(a+1,a+n+1);
  for(int i=2,j=n;i<=j;i++){
    int t=max(a[i]-a[i-1]-1,0);
    if(j-i+1<=t)ans+=j-i+1,j=i-1;
    else j-=t,ans+=t;
  }
  printf("%d\n",ans);
  return 0;
}

F题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 4005
using namespace std;
int n,f[2][maxn][180][90],a[maxn],sum[maxn];
const int inf=1e9;
int read(){
  int x=0,f=1;char ch;
  for(ch=getchar();ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
  for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
  return x*f;
}
int dfs(int z,int l,int r,int k){
  if(f[z][l][r][k]!=-1)return f[z][l][r][k];
  if(z==0)f[z][l][r][k]=max(2*l+r-90+k>n?0:dfs(1,l+k,r-k,k)+sum[l+k]-sum[l],2*l+r-90+k+1>n?-inf:dfs(1,l+k+1,r-k-1,k+1)+sum[l+k+1]-sum[l]);
  else f[z][l][r][k]=min(2*l+r-90+k>n?0:dfs(0,l,r+k,k)-sum[n-l-r+90]+sum[n-l-r+90-k],2*l+r-90+k+1>n?inf:dfs(0,l,r+k+1,k+1)-sum[n-l-r+90]+sum[n-l-r+90-k-1]);
  return f[z][l][r][k];
}
int main(){
  n=read();for(int i=1;i<=n;i++)a[i]=read(),sum[i]=sum[i-1]+a[i];
  memset(f,-1,sizeof(f));
  printf("%d\n",dfs(0,0,90,1));
  return 0;
}

  

posted @ 2016-12-13 14:41 I'mLS 阅读(...) 评论(...) 编辑 收藏