学步园

冬令时之后，这cf的时间也是醉了 - 。-||,而且这场还是unrated...

A. Factory

一个工厂第一天开始有n个物品，每天结束生产n%m件物品，则第二天开始时有n+n%m件物品，问会不会在某天结束之后物品数%m==0

#include <cstdio>                                   //如果存在k使n*2^k%m==0,则生产停止
#include <cstring>                                  //k的最大值为O(log(2)m)
						    //枚举到20天就行，如果不停止
typedef long long LL;                               //以后也不会停止

int main() {
    LL n, m;
    scanf("%lld%lld", &n, &m);
    bool ok = true;
    LL s = n;
    LL fir = n%m;
    for (int i = 0; i < 20; i++) {
        s += s%m;
        if (s%m==0) {ok = false; break;}
    }
    printf(ok? "No\n" : "Yes\n");
    return 0;
}

B. Valuable Resources

给出二维平面上一些点，求包含这些点的最小正方形的面积

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
typedef long long LL;
#define INF  (int)1e9;

int main() {
    int n;
    LL maxx = -INF;
    LL maxy = -INF;
    LL minx = INF;
    LL miny = INF;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        LL x, y;
        scanf("%lld%lld",&x, &y);
        maxx = max(maxx,x);
        minx = min(minx,x);
        maxy = max(maxy,y);
        miny = min(miny,y);
    }
    LL s = max((maxx-minx),(maxy-miny))*max((maxx-minx),(maxy-miny));
    printf("%lld\n",s);
    return 0;
}

C. Bits

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
typedef long long LL;

int main() {
    int tcase;
    scanf("%d", &tcase);
    while (tcase--) {
        LL ll, rr;
        scanf("%lld%lld", &ll, &rr);                       // 感觉这种方法比题解给的好
        LL tep = 1;                                        //为了使这个数尽量小，我们尽可能的从低位开始补1
        for (LL i = 0; i < 64; i++) {
            LL t = tep|ll;                                 
            if (t > rr) break;
            ll = t;
            tep <<= 1;
        }
        printf("%lld\n", ll);
    }
    return 0;
}

#include <cstdio>
#include <algorithm>
int A[2100000];

int main() {
    int n, tep;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &tep);
        A[tep] = tep;
    }
    for (int i = 1; i <= 2000000; i++)
        if (A[i] == 0) A[i] = A[i-1];
    int maxx = 0;
    for (int i = 1; i <= 2000000; i++) 
        if (A[i] == i) 
            for (int j = i-1; j <= 2000000; j+= i)
                if (i <= A[j] && maxx < A[j]%i) maxx = A[j]%i;
    printf("%d\n", maxx);
    return 0;
}