Codeforces Round #587 (Div. 3)

C. White Sheet

 

一道基础的几何题  给定三个矩形的左下角坐标和右上角坐标，第一块是白色矩形，第二第三都是黑色矩形，问白色矩形是否被黑色矩形完全覆盖。思路就是根据点判断分类讨论。

1、如果白色矩形的四个点没有全部被覆盖了   那么直接输出“NO”

2、如果白色矩形的四个点都被覆盖了，那么就再判断下黑色的矩形有没有交集

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <random>

#define LL long long

bool in(LL x1,LL y1,LL x2,LL y2,LL xx2,LL yy2) {
    if (x1 >= x2 && x1 <= xx2 && y1>=y2 && y1<=yy2) {
        return true;
    }
    return false;
}

int main() {
    LL x1,y1,x2,y2,x3,y3;
    LL xx1,yy1,xx2,yy2,xx3,yy3;
    std::cin >> x1 >> y1 >> xx1 >> yy1;
    std::cin >> x2 >> y2 >> xx2 >> yy2;
    std::cin >> x3 >> y3 >> xx3 >> yy3;
    if ((in(x1,y1,x2,y2,xx2,yy2) || in(x1,y1,x3,y3,xx3,yy3)) &&
       (in(x1,yy1,x2,y2,xx2,yy2) || in(x1,yy1,x3,y3,xx3,yy3)) &&
       (in(xx1,y1,x2,y2,xx2,yy2) || in(xx1,y1,x3,y3,xx3,yy3)) &&
       (in(xx1,yy1,x2,y2,xx2,yy2) || in(xx1,yy1,x3,y3,xx3,yy3))) {
        if (in(x2,y2,x3,y3,xx3,yy3) || in(x2,yy2,x3,y3,xx3,yy3) || in(xx2,y2,x3,y3,xx3,yy3) || in(xx2,yy2,x3,y3,xx3,yy3)
            || in(x3,y3,x2,y2,xx2,yy2) || in(x3,yy3,x2,y2,xx2,yy2) || in(xx3,y3,x2,y2,xx2,yy2) || in(xx3,yy3,x2,y2,xx2,yy2)) {
            std::cout << "NO" << std::endl;
        }
        else {
            if ((in(x1,y1,x2,y2,xx2,yy2) && in(x1,yy1,x2,y2,xx2,yy2) && in(xx1,y1,x2,y2,xx2,yy2) && in(xx1,yy1,x2,y2,xx2,yy2)) ||
                (in(x1,y1,x3,y3,xx3,yy3) && in(x1,yy1,x3,y3,xx3,yy3) && in(xx1,y1,x3,y3,xx3,yy3) && in(xx1,yy1,x3,y3,xx3,yy3))) {
                std::cout << "NO" << std::endl;
            } 
            else 
                std::cout << "YES" << std::endl;
        }
    }
    else {
        std::cout << "YES" << std::endl;
    }
    return 0;
}

E2. Numerical Sequence (hard version)

这道题我觉得非常有意思。

如果我们取i从1开始 那么[10^(i-1),10^i) 里面每个sum[i] = i

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <random>

#define ll long long


inline ll get_sum ( ll x ) { //从1~区间长度x的前x项和

    return x*(x+1)/2;
}

inline ll calc_1 ( ll x ) { // 获取ssum[x]

    ll ans=0, i=1, j=1;
    for ( ; j*10<=x; i++,j*=10 ) {  // 10*j == 10^i
        // 10^i 之前的  +  10^i->x 之间的
        ans += i*get_sum(j*9) + i*j*9*(x-j*10+1);  // i * [1,(10^i-10^(i-1))] == i * [1,(10*j-j)] == i * get_sum(j*9)
                                                   // 一个而言需要增加：i*j*9 == (i * (10^i-1 - (10^(i-1)-1) ) ) == (i * (10 * j - j))
                                                   // 一共有(x - 10^i + 1) == (x - 10 * j + 1)
    }
    // 加上后面部分的
    return ans + i*get_sum(x-j+1);
}

inline ll calc_2 ( ll x ) { // 获取sum[x]

    ll ans=0, i=1, j=1;
    for ( ; j*10<=x; i++,j*=10 ) {
        ans += i*j*9;
    }
    // 加上后面部分的
    return ans + i*(x-j+1);
}

int main() {

    ll q, k;
    scanf("%lld", &q);

    while (q--) {

        scanf("%lld", &k);
        ll l = 0, r = 1e9, res1, res2;

        while ( l<=r ) { // 第一次二分确定k在哪个块里

            ll mid = l+r >> 1;
            if ( calc_1(mid) < k ) {
                l = mid + 1;
                res1 = mid;
            }
            else r = mid - 1;
        }
        k -= calc_1(res1);
        l = 0, r = res1 + 1;
        while ( l<=r ) { // 第二次二分确定k在哪个数里

            ll mid = l+r >> 1;
            if ( calc_2(mid) < k ) {
                l = mid + 1;
                res2 = mid;
            }
            else r = mid - 1;
        }
        k -= calc_2(res2 ++);

        int a[30];
        memset(a, 0, sizeof(a));
        int i = 0;
        while(res2) {
            a[++i] = res2%10;
            res2 /= 10;
        }
        printf("%d
", a[i-k+1]);
    }
    return 0;
}