1008

转载来自http://blog.csdn.net/wang907553141/article/details/52294090

1008 - Fibsieve`s Fantabulous Birthday

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Time Limit: 0.5 second(s)Memory Limit: 32 MB

Fibsieve had a fantabulous (yes, it's an actual word) birthday party this year. He had so many gifts that he was actually thinking of not having a party next year.

Among these gifts there was an N x N glass chessboard that had a light in each of its cells. When the board was turned on a distinct cell would light up every second, and then go dark.

The cells would light up in the sequence shown in the diagram. Each cell is marked with the second in which it would light up.

(The numbers in the grids stand for the time when the corresponding cell lights up)

In the first second the light at cell (1, 1) would be on. And in the 5th second the cell (3, 1) would be on. Now, Fibsieve is trying to predict which cell will light up at a certain time (given in seconds). Assume that N is large enough.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case will contain an integer S (1 ≤ S ≤ 10¹⁵) which stands for the time.

Output

For each case you have to print the case number and two numbers (x, y), the column and the row number.

Sample Input

Output for Sample Input

3

8

20

25

Case 1: 2 3

Case 2: 5 4

Case 3: 1 5

 

题解：刚开始想这道题太麻烦没简化，就TLE，后来再一看，ZZ 啊刚开始就找了个对角线规律，规律都没找全面怪不得TLE

题解：蓝色的是 奇数的平方；红色的是 偶数的平方；黄色的是对角线：对角线满足规律 n * （n - 1）+ 1，其中 n 为行（列）数； 然后注意一下特殊情况就行了

嗯个人理解表示我找不到规律    嗯副对角线的规律需要稍微在理解下

就是分清楚他的平方根是奇数还是偶数

因为在偶数和奇数的排列是有些不同的  楼主讲的很好可以

觉得有点不好找出来

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define LL long longusing namespace std;LL n;int main(){int t;scanf("%d",&t);for(int o=1;o<=t;o++){    printf("Case %d: ",o);scanf("%lld",&n);LL b,a=(LL)sqrt(n);if(a*a==n) //特殊情况{if(a&1)printf("1 %lld\n",a);//是奇数的话 那么就在第一列了elseprintf("%lld 1\n",a);//是偶数的话  那么就在第一行了continue;}//这边可以理解 这个是平方的规律/*        那个数字表示平方根 假设那个数字是3 那么   a=3  b=3*4+1=13是奇数 并且11<13那么为   2,4假设那个数字是3  那么   a=13  b=15是奇数 并且15>=13 则为   4,2 假设是 4 是偶数 那么a=4 b=2124>=21   2  5假设是 4 是偶数 则为5,218<=21 那么 5,2   */if(a&1)//如果这个是数是奇数的话{b=a*(a+1)+1;if(n<b)printf("%lld %lld\n",a+1-(b-n),a+1);elseprintf("%lld %lld\n",a+1,a+1-(n-b));}else{b=a*(a+1)+1;if(n<b)printf("%lld %lld\n",a+1,a+1-(b-n));elseprintf("%lld %lld\n",a+1-(n-b),a+1);}}return 0;}