Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1722 Accepted Submission(s): 1367
题解:好菜,自己竟然没推出来。。。
总对数n^3*(n^3-1)/2;
点数为4的对数:一行有n-1对,一个面有总共n*n行,三个面面要乘以3;
总的减去4的就是<=2的了;
代码:
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define P_ printf(" ") const int INF=0x3f3f3f3f; int main(){ int n; while(~scanf("%d",&n)){ printf("%d\n",n*n*n*(n*n*n-1)/2-n*n*(n-1)*3); } return 0; }
原文:http://www.cnblogs.com/handsomecui/p/5244701.html