hdu1085】的更多相关文章

Holding Bin-Laden Captive! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23275    Accepted Submission(s): 10358 Problem Description We all know that Bin-Laden is a notorious terrorist, and he…
简单的母函数应用. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<sta…
Description We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! “Oh, God! How terrible! ” Don’t be so afraid, guys. Although he hides in a cave…
#include <iostream> #include <cstring> using namespace std; int n[3],a[9000],b[9000],i,j,k,last,last2; int v[3]={1,2,5}; int main() { while ((cin>>n[0]>>n[1]>>n[2])&&(n[0]!=0||n[1]!=0|n[2]!=0)) { a[0]=1; last=0; for (…
列出生成函数的多项式之后暴力乘即可 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=20005; int n,x,y,z,a[N],b[N]; int main() { while(scanf("%d%d%d",&x,&y,&z)&&x+y+z) { memset(a,0,sizeof(…
题意:给a个1.b个2.c个5,求不能构成最小的数 思路: 先求1能构成的所有数,2能构成的所有数,5能构成的所有数,它们的方法数显然都是1,现在考虑把3者结合在一起,由于结果为和的形式,而又是循环加的,所以考虑用多项式来表示状态,然后进行两次卷积运算就行了. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43…
1.TreeMap和TreeSet类:A - Language of FatMouse ZOJ1109B - For Fans of Statistics URAL 1613 C - Hardwood Species POJ 2418D - StationE - Web Navigation ZOJ 1061F - Argus ZOJ 2212G - Plug-in2.SegmentTreeA-敌兵布阵 hdu 1166B - I Hate It HDU 1754C - A Simple Pro…
POJ题目分类 | POJ题目分类 | HDU题目分类 | ZOJ题目分类 | SOJ题目分类 | HOJ题目分类 | FOJ题目分类 | 模拟题: POJ1006 POJ1008 POJ1013 POJ1016 POJ1017 POJ1169 POJ1298 POJ1326 POJ1350 POJ1363 POJ1676 POJ1786 POJ1791 POJ1835 POJ1970 POJ2317 POJ2325 POJ2390 POJ1012 POJ1082 POJ1099 POJ1114…
排列组合是数学中的一个分支.在计算机编程方面也有非常多的应用,主要有排列公式和组合公式.错排公式.母函数.Catalan Number(卡特兰数)等. 一.有关组合数学的公式 1.排列公式   P(n,r)=n!/r! 2.组合公式   C(n,r)=n!/(r!*(n-r)!)  C(n,r)=C(n-1,r)+C(n-1,r-1) 3.错排公式   d[1]=0;   d[2]=1; d[n]=(n-1)*(d[n-1]+d[n-2]) 4.卡特兰数 前几项:1, 2, 5, 14, 42,…