杭电hdu 5626 Clarke and points【思维】

Clarke and points

Problem Description

 

 The Manhattan Distance between point A(XA,YA) and B(XB，YB) is |XA - XB| + |Xb - YB|;

the coordinate of each point is generated by the followed code.

Input

long long seed;
inline long long rand(long long l, long long r) {
  static long long mo=1e9+7, g=78125;
  return l+((seed*=g)%=mo)%(r-l+1);
}

cin >> n >> seed;
for (int i = 0; i < n; i++)
  x[i] = rand(-1000000000, 1000000000),
  y[i] = rand(-1000000000, 1000000000);

 

Output

For each test case, print a line with an integer represented the maximum distance.

 

Sample Input

2

3 233

5 332

 

Sample Output

1557439953

1423870062

我们知道曼哈顿距离的公式、当我们计算他的时候，因为有绝对值，所以我们要考虑正负值的关系，这里一共的可能有4种：

正+正、

负+负、

正+负、

负+正、

如果这里大家对官方题解的解释不是很理解，大家可以画出|x+1|+|x-2|的图像，大家对应图像就能明白了，尽量让两个值同正或者是同负的时候值会越来越大、函数图像是一眼就能看出来滴~

蓝后捏：我们对式子进行简单处理、

同正：xa-xb+ya-yb=(xa+ya）-（xb+yb）、同负的时候同理：xb-xa+yb-ya=(xa-ya)-(xb-yb)、将两个式子推出来之后，我们需要做的事情就是维护最大最小的这么几个值、

AC代码：

#include<stdio.h>#include<ctime>#include<math.h>#include<iostream>#define LL long long intusing namespace std;LL seed;inline LL rand(LL l, LL r){    static LL mo = 1e9 + 7, g = 78125;    return l + ((seed *= g) %= mo) % (r - l + 1);}int main(){    int T, n;    LL maxnx, maxny, minnx, minny;    scanf("%d",&T);    while(T--) {        scanf("%d%I64d\n", &n, &seed);        maxnx = maxny = -0x3f3f3f3f;//这里设定要足够大才行        minnx = minny = 0x3f3f3f3f;        for(int i = 1; i <= n; ++i) {            LL x = rand(-1000000000, 1000000000);            LL y = rand(-1000000000, 1000000000);            maxnx = max(maxnx , x + y);            maxny = max(maxny , x - y);            minnx = min(minnx , x + y);            minny = min(minny , x - y);        }        printf("%I64d\n", max(maxnx - minnx, maxny - minny));    }    return 0;}