ADDED   SRM/518-U/1C.cpp
Index: SRM/518-U/1C.cpp
==================================================================
--- SRM/518-U/1C.cpp
+++ SRM/518-U/1C.cpp
@@ -0,0 +1,140 @@
+#include <iostream>
+#include <sstream>
+#include <iomanip>
+#include <vector>
+#include <string>
+#include <map>
+#include <set>
+#include <algorithm>
+#include <numeric>
+#include <iterator>
+#include <functional>
+#include <complex>
+#include <queue>
+#include <valarray>
+#include <stack>
+#include <cmath>
+#include <cassert>
+#include <cstring>
+using namespace std;
+typedef long long LL;
+typedef complex<double> CMP;
+
+static const int MODVAL = 1000000007;
+struct mint
+{
+	int val;
+	mint():val(0){}
+	mint(int    x):val(x%MODVAL) {}
+	mint(size_t x):val(x%MODVAL) {}
+	mint(LL     x):val(x%MODVAL) {}
+};
+mint& operator+=(mint& x, mint y) { return x = x.val+y.val; }
+mint& operator-=(mint& x, mint y) { return x = x.val-y.val+MODVAL; }
+mint& operator*=(mint& x, mint y) { return x = LL(x.val)*y.val; }
+mint POW(mint x, LL e) { mint v=1; for(;e;x*=x,e>>=1) if(e&1) v*=x; return v; }
+mint& operator/=(mint& x, mint y) { return x *= POW(y, MODVAL-2); }
+mint operator+(mint x, mint y) { return x+=y; }
+mint operator-(mint x, mint y) { return x-=y; }
+mint operator*(mint x, mint y) { return x*=y; }
+mint operator/(mint x, mint y) { return x/=y; }
+mint inv2 = mint(1) / 2;
+
+class Nim { public:
+	int count(int K, int L)
+	{
+		valarray<mint> v(65536);
+		v[slice(2,L-1,1)] = 1;
+		for(unsigned p=2; p<=L; ++p)
+			if( v[p].val )
+				for(unsigned q=p*p; q<=L; q+=p)
+					v[q] = 0;
+
+		pre(v, 0, 65536);
+		valarray<mint> a(1, 65536);
+		for(int i=1; i<=K; i<<=1, v*=v)
+			if( K & i )
+				a *= v;
+		post(a, 0, 65536);
+		return a[0].val;
+	}
+
+	void pre(valarray<mint>& v, int s, int e)
+	{
+		if( s+1 == e )
+			return;
+		int m = (s+e)/2;
+		pre(v, s, m);
+		pre(v, m, e);
+		for(int i=s,j=m; i<m; ++i,++j) {
+			mint vi=v[i], vj=v[j];
+			v[i] = vi - vj;
+			v[j] = vi + vj;
+		}
+	}
+
+	void post(valarray<mint>& v, int s, int e)
+	{
+		if( s+1 == e )
+			return;
+		int m = (s+e)/2;
+		for(int i=s,j=m; i<m; ++i,++j) {
+			mint dif=v[i], sum=v[j];
+			v[i] = (sum + dif) * inv2;
+			v[j] = (sum - dif) * inv2;
+		}
+		post(v, s, m);
+		post(v, m, e);
+	}
+};
+
+// BEGIN CUT HERE
+#include <ctime>
+double start_time; string timer()
+ { ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
+template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
+ { os << "{ ";
+   for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
+   os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
+void verify_case(const int& Expected, const int& Received) {
+ bool ok = (Expected == Received);
+ if(ok) cerr << "PASSED" << timer() << endl;  else { cerr << "FAILED" << timer() << endl;
+ cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
+#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
+#define END	 verify_case(_, Nim().count(K, L));}
+int main(){
+
+CASE(0)
+	int K = 3; 
+	int L = 7; 
+	int _ = 6; 
+END
+CASE(1)
+	int K = 4; 
+	int L = 13; 
+	int _ = 120; 
+END
+CASE(2)
+	int K = 10; 
+	int L = 100; 
+	int _ = 294844622; 
+END
+CASE(3)
+	int K = 123456789; 
+	int L = 12345; 
+	int _ = 235511047; 
+END
+CASE(4)
+	int K = 1000000000;
+	int L = 50000; 
+	int _ = 428193537; 
+END
+/*
+CASE(5)
+	int K = ; 
+	int L = ; 
+	int _ = ; 
+END
+*/
+}
+// END CUT HERE

ADDED   lib/doc/fft.txt
Index: lib/doc/fft.txt
==================================================================
--- lib/doc/fft.txt
+++ lib/doc/fft.txt
@@ -0,0 +1,63 @@
+
+http://apps.topcoder.com/wiki/display/tc/SRM+518
+
+
+以下のような関数を考える。
+
+int[] xor_count(int[] a, int[] b, int N)
+{
+	int[] c = {0} * N;
+	for(int i=0; i<N; ++i)
+		for(int k=0; k<N; ++k)
+			c[i ^ k] += a[i] * b[k];
+	return c;
+}
+
+高速化。
+
+int[] xor_count_ultra(int[] a, int[] b, int N)
+{
+	a = xor_pre(a);
+	b = xor_pre(b);
+	c = a * b; // element-wise product
+	return xor_post(c);
+}
+
+このようになるマジカルな変換 xor_pre, xor_post が存在する。
+
+N=1 の場合、pre : {x} --> {x}
+N=2 の場合、pre : {x,y} --> {x-y, x+y}
+  xor_count( {x,y}, {u,v} ) = {xu+yv, yu+xv}
+  と {x-y, x+y} * {u-v, u+v} = {xu+yv-xu-xv, xu+yv+yu+xv}
+  から確認できる。
+一般にNが2のベキの場合、xorの作用するビットのパターンが再帰的なので
+  pre( [X,Y] ) = [pre(X)-pre(Y), pre(X)+pre(Y)]
+と再帰的にすれば成り立つ
+
+
+
+
+
+
+xorでなくて足し算の場合
+
+int[] plus_count(int[] a, int[] b, int N)
+{
+	int[] c = {0} * 2N;
+	for(int i=0; i<N; ++i)
+		for(int k=0; k<N; ++k)
+			c[i + k] += a[i] * b[k];
+	return c;
+}
+
+も同様に。
+
+int[] plus_count_ultra(int[] a, int[] b, int N)
+{
+	a = plus_pre(a);
+	b = plus_pre(b);
+	c = a * b; // element-wise product
+	return plus_post(c);
+}
+
+とできるのが FFT