A11624. A Creative Cutout
编程题
普及/提高-
知识点
题目描述
The first line contains one integer $m$ $(1<=m<=10^{12})$ .
输入格式
In the first line print one integer representing .
输出格式
A piece of paper with $5$ circles is shown in the following.
There are $5$ types of lattice points where the degree of beauty of each red point is $1+2+3+4+5=15$ , the degree of beauty of each orange point is $2+3+4+5=14$ , the degree of beauty of each green point is $4+5=9$ , the degree of beauty of each blue point is $5$ and the degree of beauty of each gray point is $0$ . Therefore, $f(5)=5·15+4·14+4·9+8·5=207$ .
Similarly, $f(1)=5,f(2)=23,f(3)=50,f(4)=102$ and consequently .
There are $5$ types of lattice points where the degree of beauty of each red point is $1+2+3+4+5=15$ , the degree of beauty of each orange point is $2+3+4+5=14$ , the degree of beauty of each green point is $4+5=9$ , the degree of beauty of each blue point is $5$ and the degree of beauty of each gray point is $0$ . Therefore, $f(5)=5·15+4·14+4·9+8·5=207$ .
Similarly, $f(1)=5,f(2)=23,f(3)=50,f(4)=102$ and consequently .
输入输出样例
输入 #1
5
输出 #1
387
输入 #2
233
输出 #2
788243189
说明/提示
A piece of paper with $5$ circles is shown in the following.
There are $5$ types of lattice points where the degree of beauty of each red point is $1+2+3+4+5=15$ , the degree of beauty of each orange point is $2+3+4+5=14$ , the degree of beauty of each green point is $4+5=9$ , the degree of beauty of each blue point is $5$ and the degree of beauty of each gray point is $0$ . Therefore, $f(5)=5·15+4·14+4·9+8·5=207$ .
Similarly, $f(1)=5,f(2)=23,f(3)=50,f(4)=102$ and consequently .
There are $5$ types of lattice points where the degree of beauty of each red point is $1+2+3+4+5=15$ , the degree of beauty of each orange point is $2+3+4+5=14$ , the degree of beauty of each green point is $4+5=9$ , the degree of beauty of each blue point is $5$ and the degree of beauty of each gray point is $0$ . Therefore, $f(5)=5·15+4·14+4·9+8·5=207$ .
Similarly, $f(1)=5,f(2)=23,f(3)=50,f(4)=102$ and consequently .