湖南集训DAY2

水题 统计0的前缀和 

枚举一个位置i 判断i之前1的个数和i后面0的个数的和 

取小

#include <cstdio>
#include <cctype>
#include <cstring>

const int MAXN=100010;

int n,cnt,ans;

int sum[MAXN];

char s[MAXN];

int hh() {
//    freopen("reverse.in","r",stdin);
//    freopen("reverse.out","w",stdout);
    
    scanf("%s",s+1);
    n=strlen(s+1);
    
    for(int i=1;i<=n;++i) {
        if(s[i]=='0') ++cnt,sum[i]=sum[i-1]+1;
        else sum[i]=sum[i-1];
    }
    
    ans=1e9;
    for(int i=1;i<=n;++i) {
        int t=0;
        t+=i-1-sum[i-1];
        t+=sum[n]-sum[i];
        ans=ans>t?t:ans;
    }
    printf("%d
",ans);

    return 0;
}

int sb=hh();
int main(int argc,char**argv) {;}

10^7只能用O(n)的做法

枚举每一个数 求一个相当于哈希值的数 

若x,y的数码种类相同 则这个哈希值也相同 

一个哈希值的个数为C(sum,2);

#include <cstdio>
#include <cctype>
#include <iostream>

typedef long long LL;

const int MAXN=10000010;

LL ans;

int n;

int prime[10]={2,3,5,7,11,13,19,17,23,29};

int prim[10]={101,103,107,109,113,127,131,137,233,151};

LL cnt[MAXN];

bool vis[MAXN][10];

int hh() {
//    freopen("number.in","r",stdin);
//    freopen("number.out","w",stdout);
    
    scanf("%d",&n);

    int t,l,p=0;
    for(int i=1;i<=n;++i) {
        t=i;p=0;
        while(t) {
            l=t%10;
            if(!vis[i][l]) p+=prim[l]*prime[l]+l*prime[l];
            vis[i][l]=true;
            t/=10;
        }
        ++cnt[p];
    }
    for(int i=1;i<=1000000;++i) 
      if(cnt[i]>1) ans+=(LL)(cnt[i]-1)*(LL)cnt[i]/2;
    
    std::cout<<ans;
    return 0;
}

int sb=hh();
int main(int argc,char**argv) {;}

贪心 使偶数项上的数尽量大，使奇数项上的数尽量小

#include <cstdio>
#include <cctype>
#define max(a,b) a<b?b:a
#define min(a,b) a>b?b:a

const int MAXN=2000010;

int n,k,ans;

int a[MAXN];

inline void read(int&x) {
    int f=1;register char c=getchar();
    for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
    for(;isdigit(c);x=x*10+c-48,c=getchar());
    x=x*f;
}

int hh() {
    freopen("wave.in","r",stdin);
    freopen("wave.out","w",stdout);

    read(n);read(k);
    for(int i=1;i<=n;++i) read(a[i]);
    
    int last=a[1],flag=1,ans=1;
    for(int i=2;i<=n;++i) {
        if(flag) {
            if(a[i]>=last+k) flag^=1,last=a[i],++ans;
            else last=min(last,a[i]);
        }
        else {
            if(last>=a[i]+k) flag^=1,last=a[i],++ans;
            else last=max(last,a[i]);
        }
    }
    printf("%d
",ans);
    
    return 0;
}

int sb=hh();
int main(int argc,char**argv) {;}