2017-10-17 NOIP模拟赛

Reverse

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char c[];
int len,cnt,sum[];
struct node{
    int n1,n0;
}a[];
int main(){
    freopen("reverse.in","r",stdin);freopen("reverse.out","w",stdout);
    //freopen("Cola.txt","r",stdin);
    scanf("%s",c+);
    len=strlen(c+);
    int x1=,x0=;
    for(int i=;i<=len;i++){
        if(c[i]=='')x0++;
        if(c[i]=='')x1++;
        if((c[i]==''&&c[i+]=='')||i==len){
            a[++cnt].n0=x0;
            a[cnt].n1=x1;
            x0=;x1=;
        }
    }
    for(int i=cnt;i>=;i--)sum[i]=sum[i+]+a[i].n0;
    int ans=;
    for(int i=;i<=cnt;i++){
        if(a[i].n1<sum[i+])ans+=a[i].n1;
        else{
            ans+=sum[i+];
            break;
        }
    }
    printf("%d",ans);
    return ;
}

50分 迷之wa

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 100100
using namespace std;
int len,ans=0x7f7f7f7f;
char s[MAXN];
int sum[MAXN];
int main(){
    freopen("reverse.in","r",stdin);
    freopen("reverse.out","w",stdout);
    scanf("%s",s);
    len=strlen(s);
    sum[]=s[]-'';
    for(int i=;i<len;i++)
        sum[i]=sum[i-]+s[i]-'';
    for(int i=;i<len;i++)
        ans=min(ans,sum[i]+(len--i-sum[len-]+sum[i]));
    cout<<ans;
}  

100分 前缀和+O(n)扫一遍

Number

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#ifdef WIN32
#define PLL "%I64d"
#else
#define PLL "%lld"
#endif
using namespace std;
int n,cnt[];
long long ans;
short b[];
int count(int x){
    memset(cnt,,sizeof(cnt));
    while(x){
        cnt[x%]++;
        x/=;
    }
    int res=;
    for(int i=;i>=;i--){
        if(cnt[i])res=res*+i;
    }
    return res;
}
int main(){
    //freopen("Cola.txt","r",stdin);
    freopen("number.in","r",stdin);freopen("number.out","w",stdout);
    scanf("%d",&n);
    for(int i=;i<=n;i++){
        int now=count(i);
        ans+=b[now];
        b[now]++;
    }
    printf(PLL,ans);
    return ;
}

75分 桶

/*
枚举，用vis一个二进制数组表示当前数每一位的状态，出现或者不出现。
然后把二进制转为十进制，cnt[十进制]+1。
最后因为要求多少对，所以答案累加C(i,2)。
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define size 1024

using namespace std;
int n,cnt[size],vis[];
int Use[]={,,,,,,,,,};
long long ans;

int main()
{
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    cin>>n;
    int start,End;
    for(int i=;i<=n;i++)
    {
        start=i;End=;memset(vis,,sizeof vis);
        while(start) vis[start%]=,start/=;
        for(int j=;j<=;j++) if(vis[j]) End+=Use[j];
        cnt[End]++;
    }
    for(int i=;i<size;i++) ans+=1ll*cnt[i]*(cnt[i]-)/;
    cout<<ans<<endl;
    fclose(stdin);fclose(stdout);
    return ;
}

100分 桶

Wave

#include<iostream>
#include<cstdio>
#define maxn 2000010
using namespace std;
int f[maxn][],n,k,a[maxn],ans;
int qread(){
    int i=,j=;
    char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')j=-;ch=getchar();}
    while(ch<=''&&ch>=''){i=i*+ch-'';ch=getchar();}
    return i*j;
}
int main(){
    freopen("wave.in","r",stdin);freopen("wave.out","w",stdout);
    //freopen("Cola.txt","r",stdin);
    n=qread();k=qread();
    for(int i=;i<=n;i++){
        a[i]=qread();
        f[i][]=;
    }
    for(int i=;i<=n;i++){
        for(int j=;j<i;j++){
            if(a[j]-a[i]>=k){//i是奇数项
                f[i][]=max(f[i][],f[j][]+);
                ans=max(ans,f[i][]);
            }
            if(a[i]-a[j]>=k){//i是偶数项
                f[i][]=max(f[i][],f[j][]+);
                ans=max(ans,f[i][]);
            }
        }
    }
    printf("%d",ans);
}

20分 暴力dp

/*
很神奇。
贪心，在满足k的限制下，偶数项尽量大，奇数项尽量小。
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 2000007

using namespace std;
int a[N];
int n,k,m,ans,cur;

inline int read()
{
    int x=,f=;char c=getchar();
    while(c>''||c<''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

int main()
{
    n=read();k=read();
    for(int i=;i<n;i++) a[i]=read();
    ans=,m=,cur=a[];
    for(int i=;i<n;i++)
    {
        if(m)
          if(cur-a[i]>=k)
            m=,ans++,cur=a[i];
          else cur=max(cur,a[i]);

        else
          if(a[i]-cur>=k)
            m=,ans++,cur=a[i];
          else cur=min(cur,a[i]);
    }
    printf("%d\n",ans);
    return ;
}

100分 贪心