Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem J

(2/3/4)-DSqr/Rects/Cubes/Boxes?

Input: standard input

Output: standardoutput

Time Limit:2 seconds

 

You can see a (4x4) grid below.Can you tell me how many squares and rectangles are hidden there? You can assumethat squares are not rectangles. Perhaps one can count it by hand but can youcount it for a (100x100) grid or a (10000x10000) grid. Can you do it for higherdimensions? That is can you count how many cubes or boxes of different size arethere in a (10x10x10) sized cube or how many hyper-cubes or hyper-boxes ofdifferent size are there in a four-dimensional (5x5x5x5) sized hypercube.Remember that your program needs to be very efficient. You can assume thatsquares are not rectangles, cubes are not boxes and hyper-cubes are nothyper-boxes. 

 

 

 

Input

The input contains one integer N(0<=N<=100) in each line, which is the length of one side of the grid or cubeor hypercube. As for the example above the value of N is 4. There may be asmany as 100 lines of input.

 

Output

For each line of input, output six integers S2, R2, S3,R3, S4, R4 in a single line where S2 means no of squares of different size in (NxN) two-dimensional grid, R2 means no of rectangles of different size in (NxN)two-dimensional grid. S3, R3, S4, R4 means similar cases in higher dimensionsas described before.  

 

Sample Input:

1
2
3

Sample Output:

1 0 1 0 1 0
5 4 9 18 17 64
14 22 36 180 98 1198

---

#include<stdio.h>
int main() {
    long long n;
    while(scanf("%lld", &n) == 1) {
        long long a, b, c, d, e = 0, f, i;
        a = n*(n+1)*(2*n+1)/6;
        b = (n+1)*n/2*(n+1)*n/2 - a;
        c = n*(n+1)/2*n*(n+1)/2;
        d = (n+1)*n/2*(n+1)*n/2*(n+1)*n/2 - c;
        for(i = 1; i <= n; i++)    e += i*i*i*i;
        f = (n+1)*n/2*(n+1)*n/2*(n+1)*n/2*(n+1)*n/2 - e;
        printf("%lld %lld %lld %lld %lld %lld\n", a, b, c, d, e, f);
    }
    return 0;
}