Description

I have a set of super poker cards, consisting of an infinite number of cards. For each positive composite integer p, there
are exactly four cards whose value is p: Spade(S), Heart(H), Club(C) and
Diamond(D). There are no cards of other values.
By “composite integer”, we
mean integers that have more than 2 divisors. For example, 6 is a composite
integer, since it
has 4 divisors: 1, 2, 3, 6; 7 is not a composite number,
since 7 only has 2 divisors: 1 and 7. Note that 1 is not composite
(it has
only 1 divisor).
 
Given a positive integer n, how many ways can you pick
up exactly one card from each suit (i.e. exactly one spade card,
one heart
card, one club card and one diamond card), so that the card values sum to n? For
example, if n=24, one way is
4S+6H+4C+10D, shown below:

Unfortunately, some of the cards are lost,
but this makes the problem more interesting. To further make the problem even

more interesting (and challenging!), I’ll give you two other positive
integers a and b, and you need to find out all the
answers for n=a, n=a+1,
…, n=b.

Input

The input contains at most 25 test cases.
Each test case begins with 3 integers a, b and c, where c is the number of lost

cards. The next line contains c strings, representing the lost cards. Each
card is formatted as valueS, valueH, valueC or
valueD, where value is a
composite integer. No two lost cards are the same. The input is terminated by
a=b=c=0. There
will be at most one test case where a=1, b=50,000 and
c<=10,000. For other test cases, 1<=a<=b<=100,
0<=c<=10.

Output

For each test case, print b-a+1 integers, one
in each line. Since the numbers might be large, you should output each

integer modulo 1,000,000. Print a blank line after each test
case.

Sample Input

12 20 2
4S 6H
0 0 0

Sample Output

0
0
0
0
0
0
1

0
3

HINT

很简单的fft,看懂题面即可。

code:

 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 131075
#define pi 3.14159265358979323846
#define mod 1000000
using namespace std;
typedef long long int64;
char ch;
int l,r,m,n,x,len,tot,re[maxn],prime[maxn];
bool ok,bo[maxn];
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
int rev(int v){
int t=;
for (int i=;i<len;i++) t<<=,t|=v&,v>>=;
return t;
}
struct comp{
double rea,ima;
void clear(){rea=ima=;}
comp operator +(const comp &x){return (comp){rea+x.rea,ima+x.ima};}
comp operator -(const comp &x){return (comp){rea-x.rea,ima-x.ima};}
comp operator *(const comp &x){return (comp){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}
}a[maxn],b[maxn],c[maxn],d[maxn],Wn[][maxn],wn,w,t1,t2;
void fft(comp *a,int op){
for (int i=,t=re[i];i<n;i++,t=re[i]) if (i<t) swap(a[i],a[t]);
for (int s=;s<=n;s<<=){
wn=Wn[op][s];//cout<<wn.rea<<' '<<wn.ima<<endl;
for (int i=;i<n;i+=s){
w=(comp){,};
for (int j=i;j<i+(s>>);j++,w=w*wn){
t1=a[j],t2=w*a[j+(s>>)];
a[j]=t1+t2,a[j+(s>>)]=t1-t2;
}
}
}
if (op) for (int i=;i<n;i++) a[i].rea/=n,a[i].ima/=n;
}
void work(){
for (int i=;i<=r;i++) a[i].rea=(int64)round(a[i].rea)%mod,a[i].ima=;
for (int i=r+;i<n;i++) a[i].clear();
}
void init(){
for (int i=;i<maxn;i<<=) Wn[][i]=(comp){cos(*pi/i),sin(*pi/i)};
for (int i=;i<maxn;i<<=) Wn[][i]=(comp){cos(-*pi/i),sin(-*pi/i)};
for (int i=;i<=;i++){
if (!bo[i]) prime[++tot]=i;
for (int j=;j<=tot&&i*prime[j]<=;j++){
bo[i*prime[j]]=;
if (!(i%prime[j])) break;
}
}
}
int main(){
for (init(),read(l),read(r),read(m);l&&r;read(l),read(r),read(m)){
for (len=,n=;n<((r+)<<);len++,n<<=);
for (int i=;i<n;i++) re[i]=rev(i);
for (int i=;i<n;i++) a[i].clear(),b[i].clear(),c[i].clear(),d[i].clear();
for (int i=;i<r;i++) a[i].rea=b[i].rea=c[i].rea=d[i].rea=bo[i];
for (int i=;i<=m;i++){
read(x);
if (ch=='S') a[x].rea=;
else if (ch=='H') b[x].rea=;
else if (ch=='C') c[x].rea=;
else if (ch=='D') d[x].rea=;
}
fft(a,),fft(b,),fft(c,),fft(d,);
for (int i=;i<n;i++) a[i]=a[i]*b[i];
fft(a,),work(),fft(a,);
for (int i=;i<n;i++) a[i]=a[i]*c[i];
fft(a,),work(),fft(a,);
for (int i=;i<n;i++) a[i]=a[i]*d[i];
fft(a,),work();
for (int i=l;i<=r;i++) printf("%d\n",(int)a[i].rea);
puts("");
}
return ;
}

bzoj2487: Super Poker II的更多相关文章

  1. UVA - 12298 Super Poker II NTT

    UVA - 12298 Super Poker II NTT 链接 Vjudge 思路 暴力开个桶,然后统计,不过会T,用ntt或者fft,ntt用个大模数就行了,百度搜索"NTT大模数&q ...

  2. UVa12298 Super Poker II(母函数 + FFT)

    题目 Source http://acm.hust.edu.cn/vjudge/problem/23590 Description I have a set of super poker cards, ...

  3. Super Poker II UVA - 12298 FFT_生成函数

    Code: #include<bits/stdc++.h> #define maxn 1000000 #define ll long long #define double long do ...

  4. FFT(快速傅里叶变换):UVAoj 12298 - Super Poker II

    题目:就是现在有一堆扑克里面的牌有无数张, 每种合数的牌有4中不同花色各一张(0, 1都不是合数), 没有质数或者大小是0或者1的牌现在这堆牌中缺失了其中的 c 张牌, 告诉你a, b, c接下来c张 ...

  5. UVA12298 Super Poker II

    怎么又是没人写题解的UVA好题,个人感觉应该是生成函数的大板子题了. 直接做肯定爆炸,考虑来一发优化,我们记一个多项式,其中\(i\)次项的系数就表示对于\(i\)这个数有多少种表示方式. 那么很明显 ...

  6. UVA - 12298 Super Poker II (FFT+母函数)

    题意:有四种花色的牌,每种花色的牌中只能使用数值的约数个数大于2的牌.现在遗失了c张牌.每种花色选一张,求值在区间[a,b]的每个数值的选择方法有多少. 分析:约数个数大于2,即合数.所以先预处理出5 ...

  7. UVA 12298 Super Poker II (FFT)

    #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using ...

  8. 浅谈FFT(快速傅里叶变换)

    本文主要简单写写自己在算法竞赛中学习FFT的经历以及一些自己的理解和想法. FFT的介绍以及入门就不赘述了,网上有许多相关的资料,入门的话推荐这篇博客:FFT(最详细最通俗的入门手册),里面介绍得很详 ...

  9. bzoj AC倒序

    Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem ...

随机推荐

  1. 【模拟赛】BYVoid魔兽世界模拟赛 解题报告

    题目名称(点击进入相关题解) 血色先锋军 灵魂分流药剂 地铁重组 埃雷萨拉斯寻宝 源文件名(.c/.cpp/.pas) scarlet soultap subway eldrethalas 输入文件名 ...

  2. winscp自动执行脚本

    我们经常使用WinSCP工具通过sftp协议上传获取文件,本文描述通过bat批量处理文件. 首先,我们打开dos命令窗口使用 cd \d :D\WinSCP 打开WinSCP安装目录 上传文件: wi ...

  3. poj1013

    题目大意:假造的银币 Sally Jones有一些游客给的银币,但是只有11枚是真正的银币(有一枚是假的),从颜色和大小是无法区分真比还是假币的,但是它的重量和真币是不同的,Sally Jones它是 ...

  4. Java面试求职之==与equals()差别

    Java中equals和==的差别 java中的数据类型,可分为两类:     1.基本数据类型(也称原始数据类型):byte,short,char,int,long,float,double,boo ...

  5. linux安装tomcat(转载:http://blog.csdn.net/zhuihunmiling/article/details/8977387)

    在安装Tomcat之前需要安装j2sdk(Java 2 Software Development Kit),也就是JDK 1.安装JDK完毕. 2.安装Tomcat 1)下载apache-tomcat ...

  6. php 链式操作的实现 学习记录

    php 面向对象中实现链式操作的关键部分:调用的方法中返回当前对象 ,从而实现链式操作: <?php namespace commom; class db { public function w ...

  7. Linux开发工具之gdb(上)

    三.gdb调试(上) 01.gdb:gdb是GNU debugger的缩写,是编程调试工作. 功能:   启动程序,可以按照用户自定义的要求随心所欲的运行程序:   可让被调试的程序在用户所指定的调试 ...

  8. ubuntu wine卸载程序并删除图标

    卸载ubuntu 下用wine安装的程序,可以用wine uninstaller命令,打开 添加/删除程序界面,进行删除程序操作:

  9. post 封装Map 发送请求

    package com.j1.weixin.util; import java.io.IOException; import java.util.Map; import java.util.Set; ...

  10. 【开源java游戏框架libgdx专题】-11-核心库-演员类

    演员类,又称为Actor类,是libgdx开发中最基本的元素,可以被继承. 演员类,从OpenGL类的角度来理解,可以称为一个二维场景节点.它本身具有位置(postion).边界矩形(类似Retang ...