• JNday3-am


    预计分数 ? + 0 + 50;

    实际分数 A + 0 + 50;

    这套题感觉还行

    T1感觉没啥思路,首先放弃,先做T3

    T3的思路之前见过,然后没多长时间就写了出来,然后自己调试了很长时间过了样例,
    之后由于一些不可描述的原因,GG了,(这个原因我也不想再说了。。。。。。)
    然后写了个暴力 + 一个线段树的区间查询最大值的板子题;

    T1看了看样例,每组数据只要有偶数,那么就可以,虽然不能严谨的证明,但是骗过了

    现在,已经没多长时间了。。。。。。(貌似还有不到15min)

    T2表面一看,题面和昨天的T2一样,貌似是一个30min可以写出的爆搜,然而时间不够,就放弃了,
    然后看到很多40分的人,就读了一下题面,原来一个简单的 N^2 就可以过,GG,竟然没看,丢了40分

    总结:T2凭经验放弃,这是很不理智的,(起码得看一下吧)
    T1骗分比较成功(其实yy出来的就是正解嘛)
    T3 板子的东西一定要敲的非常熟练,如果T3敲的快,那就还有不少时间来“看”T2吧。

     T1 星空

    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    
    using namespace std;
    
    int n,m;
    
    int main()
    {
        freopen("star.in","r",stdin);
        freopen("star.out","w",stdout);
    
        while (~scanf("%d%d",&n,&m))
        {
            if (!n && !m) break;
            if ((n&1)&&(m&1)) printf("Chito
    ");
            else printf("Yuri
    ");
        }
    
        return 0;
    }

    T2 战争

    #define PROC "shana"
    #include <cstdio>
    #include <cctype>
    #include <memory.h>
    #include <algorithm>
    #include<cctype>
    
    using namespace std;
    
    typedef long long qw;
    
    #define _l (qw)
    const int BUF_SIZE = 30;
    char buf[BUF_SIZE], *buf_s = buf, *buf_t = buf + 1;
      
    #define PTR_NEXT() 
        { 
            buf_s ++; 
            if (buf_s == buf_t) 
            { 
                buf_s = buf; 
                buf_t = buf + fread(buf, 1, BUF_SIZE, stdin); 
            } 
        }
       
    #define readInt(_n_) 
        { 
            while (*buf_s != '-' && !isdigit(*buf_s)) 
                PTR_NEXT(); 
            bool register _nega_ = false; 
            if (*buf_s == '-') 
            { 
                _nega_ = true; 
                PTR_NEXT(); 
            } 
            int register _x_ = 0; 
            while (isdigit(*buf_s)) 
            { 
                _x_ = _x_ * 10 + *buf_s - '0'; 
                PTR_NEXT(); 
            } 
            if (_nega_) 
                _x_ = -_x_; 
            (_n_) = (_x_); 
        }
    
    const int maxn = 50010;
    const int maxl = 31;
    const int maxnd = maxn * maxl;
    const int mod = 1e9 + 7;
    const int inv = 500000004;
    
    int v0, n, rt, tn, a[maxn];
    int tr[maxnd][2], rb[maxnd][maxl], c[maxnd];
    qw k;
    
    void trieIns(int v) {
        int p = rt;
        for (int i = maxl - 1; i >= 0; -- i) {
            int v0 = (v >> i) & 1;
            if (!tr[p][v0])
                tr[p][v0] = ++ tn;
            p = tr[p][v0];
            ++ c[p];
            for (int j = maxl - 1; j >= 0; -- j)
                if ((v >> j) & 1)
                    ++ rb[p][j];
        }
    }
    
    int cntUpper(int v, int vu) {
        int p = rt, s = 0;
        for (int i = maxl - 1; i >= 0; -- i) {
            int v0 = (v >> i) & 1;
            if ((vu >> i) & 1) {
                p = tr[p][v0 ^ 1];
            }
            else {
                s += c[tr[p][v0 ^ 1]];
                p = tr[p][v0];
            }
        }
        return s;
    }
    
    qw check(int v) {
        qw s = 0;
        for (int i = 0; i < n; ++ i)
            s += cntUpper(a[i], v);
        return s >> 1;
    }
    
    int sumUpper(int v, int vu) {
        int s = 0, p = rt;
        for (int i = maxl - 1; i >= 0; -- i) {
            int v0 = (v >> i) & 1;
            if ((vu >> i) & 1)
                p = tr[p][v0 ^ 1];
            else {
                for (int j = 0; j < maxl; ++ j)
                    if ((v >> j) & 1)
                        s = (_l s + (1LL << j) * (_l c[tr[p][v0 ^ 1]] - _l rb[tr[p][v0 ^ 1]][j])) % mod;
                    else
                        s = (_l s + (1LL << j) * _l rb[tr[p][v0 ^ 1]][j]) % mod;
                p = tr[p][v0];
            }
        }
        return s;
    }
    
    int main() {
        freopen("war.in", "r", stdin);
        freopen("war.out", "w", stdout);
    
        readInt(n);
        readInt(k);
        rt = 1;
        tn = 1;
        for (int i = 0; i < n; ++ i) {
            readInt(a[i]);
            trieIns(a[i]);
        }
        {
            int l = 0, r = 2147483647;
            while (l < r) {
                int mid = (_l l + r + 1) >> 1;
                if (check(mid - 1) < k)
                    r = mid - 1;
                else
                    l = mid;
            }
            v0 = l;
        }
        if (v0) {
            //printf("%d %lld
    ", v0, check(v0));
            int ans = 0;
            for (int i = 0; i < n; ++ i)
                ans = (_l ans + sumUpper(a[i], v0 - 1)) % mod;
            ans = (_l ans * inv % mod + ((k - check(v0 - 1)) % mod + mod) * _l v0) % mod;
            printf("%d
    ", ans);
        }
        else {
            int ans = 0;
            for (int i = 0; i < n; ++ i)
                ans = (_l ans + sumUpper(a[i], 0)) % mod;
            ans = _l ans * inv % mod;
            printf("%d
    ", ans);
        }
    }

    T3 无题

    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    
    using namespace std;
    
    #define wmt 1,n,1
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    
    const int maxn=100010;
    const int maxk=10;
    
    int n,m,col[maxn<<2|1];
    
    struct rec
    {
        int res[maxk];
        rec(){memset(res,0,sizeof(res));}
        rec operator+(const rec &b)const {
            rec z;
            int p1=0,p2=0;
            for (int a=0;a<maxk;a++)
                if (res[p1]>b.res[p2]) z.res[a]=res[p1++];
                else z.res[a]=b.res[p2++];
            return z;
        }
    }z[maxn<<2|1];
    
    void color(int rt,int c)
    {
        col[rt]+=c;
        for (int a=0;a<maxk;a++)
            if (z[rt].res[a]) z[rt].res[a]+=c;
    }
    
    void push_col(int rt) {
        if (col[rt]) {
            color(rt<<1,col[rt]);
            color(rt<<1|1,col[rt]);
            col[rt]=0;
        }
    }
    
    void update(int rt) {
        z[rt]=z[rt<<1]+z[rt<<1|1];
    }
    
    void build(int l,int r,int rt)
    {
        if (l==r) {
            scanf("%d",&z[rt].res[0]);
            return;
        }
        int m=(l+r)>>1;
        build(lson);
        build(rson);
        update(rt);
    }
    
    rec query(int l,int r,int rt,int nowl,int nowr)
    {
        if (nowl<=l && r<=nowr) return z[rt];
        push_col(rt);
        int m=(l+r)>>1;
        if (nowl<=m)
        {
            if (m<nowr) return query(lson,nowl,nowr)+query(rson,nowl,nowr);
            else return query(lson,nowl,nowr);
        }
        else return query(rson,nowl,nowr);
    }
    
    void modify(int l,int r,int rt,int nowl,int nowr,int v)
    {
        if (nowl<=l && r<=nowr)
        {
            color(rt,v);
            return;
        }
        push_col(rt);
        int m=(l+r)>>1;
        if (nowl<=m) modify(lson,nowl,nowr,v);
        if (m<nowr) modify(rson,nowl,nowr,v);
        update(rt);
    }
    
    int main()
    {
        freopen("noname.in","r",stdin);
        freopen("noname.out","w",stdout);
    
        scanf("%d%d",&n,&m);
        build(wmt);
        for (int a=1;a<=m;a++)
        {
            int opt,l,r,v;
            scanf("%d%d%d%d",&opt,&l,&r,&v);
            if (opt==0) 
            {
                if (r-l+1<v) printf("-1
    ");
                else printf("%d
    ",query(wmt,l,r).res[v-1]);
            }
            else modify(wmt,l,r,v);
        }
    
        return 0;
    }
  • 相关阅读:
    对实时的视屏流进行处理
    opencv读取并播放avi视屏
    opencv中读取显示图像
    为什么既要有IP地址还要有MAC地址
    继承中构造、析构 与 拷贝构造、赋值中的调用区别
    拷贝构造函数和赋值符函数——转
    inline修饰虚函数的问题
    菱形继承产生的问题及解决
    迭代器模式
    备忘录模式
  • 原文地址:https://www.cnblogs.com/lyqlyq/p/7756822.html
Copyright © 2020-2023  润新知