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Description:
上一篇blog.
Solution:
同样我们可以用fft来做...就像上次写的那道3-idoit一样,对a做k次卷积就好了.
同样有许多需要注意的地方:我们只是判断可行性,所以为了保证精度如果f大于1就把它变成1; 对于长度也可以慢慢倍增,可以优化复杂度就是写起来麻烦.
void change(complex y[],int len)
{
int i,j,k;
for(i = 1, j = len/2;i < len-1; i++)
{
if(i < j)swap(y[i],y[j]);
k = len/2;
while( j >= k)
{
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
void fft(complex y[],int len,int on)
{
change(y,len);
for(int h = 2; h <= len; h <<= 1)
{
complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
for(int j = 0;j < len;j+=h)
{
complex w(1,0);
for(int k = j;k < j+h/2;k++)
{
complex u = y[k];
complex t = w*y[k+h/2];
y[k] = u+t;
y[k+h/2] = u-t;
w = w*wn;
}
}
}
if(on == -1)
for(int i = 0;i < len;i++)
y[i].r /= len;
}
const int maxn = 2e6+5;
complex x1[maxn], x2[maxn];
int a[maxn], b[maxn];
void cal(int *a, int *b, int &lena, int &lenb) {
int len = 1;
while(len<lena+lenb)
len<<=1;
for(int i = 0; i<=lenb; i++) {
x1[i] = complex(b[i], 0);
}
for(int i = lenb+1; i<len; i++)
x1[i] = complex(0, 0);
for(int i = 0; i<=lena; i++) {
x2[i] = complex(a[i], 0);
}
for(int i = lena+1; i<len; i++)
x2[i] = complex(0, 0);
fft(x1, len, 1);
fft(x2, len, 1);
for(int i = 0; i<len; i++)
x1[i] = x1[i]*x2[i];
fft(x1, len, -1);
for(int i = 0; i<=lena+lenb; i++)
b[i] = (int)(x1[i].r+0.5);
for(int i = 0; i<=lena+lenb; i++)
if(b[i]>0)
b[i] = 1;
lenb += lena;
}
int main()
{
int n, k, x;
cin>>n>>k;
for(int i = 0; i<n; i++) {
scanf("%d", &x);
a[x]++;
}
b[0] = 1;
int lena = 1000, lenb = 0;
while(k) {
if(k&1) {
cal(a, b, lena, lenb);
}
if(k>1) {
cal(a, a, lena, lena);
}
k>>=1;
}
for(int i = 0; i<=lena+lenb; i++) {
if(b[i]) {
printf("%d ", i);
}
}
cout<<endl;
return 0;
}
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原文地址：http://www.cnblogs.com/YCuangWhen/p/5407601.html
