FFT模板+题目

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e7+10;
const double Pi=acos(-1.0);
struct cp{
    
    
    double x,y;
    cp (double xx=0,double yy=0){
    
    x=xx,y=yy;}
}a[maxn],b[maxn];
cp operator + (cp a,cp b){
    
     return cp(a.x+b.x , a.y+b.y);}
cp operator - (cp a,cp b){
    
     return cp(a.x-b.x , a.y-b.y);}
cp operator * (cp a,cp b){
    
     return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}

int n,m,r[maxn];
void FFT(cp *A,int type, int n){
    
    
    for(int i = 0;i < n; i++)
        if(i < r[i]) swap(A[i], A[r[i]]);
    for(int mid = 1;mid < n; mid <<= 1){
    
    
        cp Wn(cos(Pi / mid) , type * sin(Pi / mid));
        for(int R = mid << 1,j = 0;j < n;j += R){
    
    
            cp w(1,0);
            for(int k = 0;k < mid;k++,w = w*Wn){
    
    
                 cp x = A[j + k],y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + mid + k]=x - y;
            }
        }
    }
}

void BU(int sum, int &n){
    
    
    int l = 0;n = 1;
    while(n <= sum)l++, n <<= 1;
    for(int i=0;i<n;i++)
        r[i]= ( r[i>>1]>>1 )| ( (i&1)<<(l-1) ) ;
}
int main(){
    
    
	//数组是从0开始的
	//BU第一个参数为卷积后的最高次幂
    int lim;
    scanf("%d %d",&n,&m);
    for(int i = 0; i <= n; i++) scanf("%lf",&a[i].x);
    for(int i = 0; i <= m; i++) scanf("%lf",&b[i].x);
    BU(n + m,lim);
    FFT(a, 1, lim),FFT(b, 1, lim);
    for(int i = 0; i <= lim; i++)a[i] = b[i] * a[i];
    FFT(a, -1, lim);

    for(int i = 0; i <= n + m; i++)printf("%d ",int(a[i].x / lim + 0.5));
    return 0;
}

HDU4609
思路：
首先先把所有不同长度的木棍进行数量统计。
然后对这个统计的数组自身进行一次卷积，那么得到的结果就是从原本的数组中的任意两个木棍进行组合的结果。
枚举每根木棍，假设它是最长的
最后只需要减去自身与自身的组合，和自身和别人组合，比自己打的木棍的组合就可以了

#include<bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1e6 + 5;
const double Pi=acos(-1.0);
struct cp{
    
    
    double x,y;
    cp (double xx=0,double yy=0){
    
    x=xx,y=yy;}
}a[maxn];
cp operator + (cp a,cp b){
    
     return cp(a.x+b.x , a.y+b.y);}
cp operator - (cp a,cp b){
    
     return cp(a.x-b.x , a.y-b.y);}
cp operator * (cp a,cp b){
    
     return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}

int n,m,r[maxn],t,y,p[maxn],mp[maxn];
ll sum[maxn];


void FFT(cp *A,int type, int n){
    
    
    for(int i=0;i<n;i++)
        if(i<r[i]) swap(A[i],A[r[i]]);
    for(int mid=1;mid<n;mid<<=1){
    
    
        cp Wn( cos(Pi/mid) , type*sin(Pi/mid));
        for(int R=mid<<1,j=0;j<n;j+=R){
    
    
            cp w(1,0);
            for(int k=0;k<mid;k++,w=w*Wn){
    
    
                 cp x=A[j+k],y=w*A[j+mid+k];
                A[j+k]=x+y;
                A[j+mid+k]=x-y;
            }
        }
    }
}

void BU(int sum, int &n){
    
    
    int l = 0;n = 1;
    while(n <= sum)l++, n <<= 1;
    for(int i=0;i<n;i++)
        r[i]= ( r[i>>1]>>1 )| ( (i&1)<<(l-1) ) ;
}

int main(){
    
    
    scanf("%d",&t);
    while(t--){
    
    
        int lim,maxx = -1;
        ll ans = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)scanf("%d",p + i),maxx = max(maxx, p[i]);
        for(int i = 1; i <= maxx; i++)mp[i] = 0;
        BU(maxx * 2, lim);
        for(int i = 0; i < n; i++)mp[p[i]]++;
        for(int i = 0; i <= maxx; i++)a[i].x = mp[i],a[i].y = 0;
        for(int i = maxx + 1; i <= lim; i++)a[i].x = a[i].y = 0;
        FFT(a,1,lim);
        for(int i = 0; i <= lim; i++)a[i] = a[i] * a[i];
        FFT(a,-1,lim);

        for(int i = 1; i <= maxx * 2; i++)sum[i] = (ll)(a[i].x / lim + 0.5);
        //for(int i = 0; i <= maxx * 2; i++)cout << sum[i] << endl;
        for(int i = 0; i < n; i++)sum[p[i] * 2] -= 1;
        for(int i = 1; i <= maxx * 2; i++)sum[i] = sum[i - 1] + sum[i] / 2;
        sort(p, p + n);

        for(int i = 0; i < n; i++){
    
    
            ans += sum[maxx * 2] - sum[p[i]];
            ans -= (n - 1);
            ans -= 1ll * (n - i - 1) * i;
            ans -= 1ll * (n - i - 1) * (n - i - 2) / 2;
        }
        ll f = 1ll * n * (n - 1) * (n - 2) / 6;
        printf("%.7lf\n",(double)ans / f);
    }
    return 0;
}

落谷
落谷题解写的很详细，这里记一下我觉着重要的部分
第一是：
$\sum\limits_{i = j}^nf[i] * g[i - j] = \sum\limits_{i = 0}^{n - j}f[j +i]*g[i]$
第二个是：
做 $f^{'} [i] = f [n - i]$ 的变换，凑出两个自变量相加等于一个常数的形式

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;


const double Pi=acos(-1.0);
struct cp{
    
    
    double x,y;
    cp (double xx=0,double yy=0){
    
    x=xx,y=yy;}
}a[maxn],b[maxn],c[maxn];
cp operator + (cp a,cp b){
    
     return cp(a.x+b.x , a.y+b.y);}
cp operator - (cp a,cp b){
    
     return cp(a.x-b.x , a.y-b.y);}
cp operator * (cp a,cp b){
    
     return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}

int n,m,r[maxn];
double f[maxn],g[maxn];
void FFT(cp *A,int type, int n){
    
    
    for(int i = 0;i < n; i++)
        if(i < r[i]) swap(A[i], A[r[i]]);
    for(int mid = 1;mid < n; mid <<= 1){
    
    
        cp Wn(cos(Pi / mid) , type * sin(Pi / mid));
        for(int R = mid << 1,j = 0;j < n;j += R){
    
    
            cp w(1,0);
            for(int k = 0;k < mid;k++,w = w*Wn){
    
    
                 cp x = A[j + k],y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + mid + k]=x - y;
            }
        }
    }
}

void BU(int sum, int &n){
    
    
    int l = 0;n = 1;
    while(n <= sum)l++, n <<= 1;
    for(int i=0;i<n;i++)
        r[i]= ( r[i>>1]>>1 )| ( (i&1)<<(l-1) ) ;
}
int main(){
    
    
    int lim;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        scanf("%lf",&a[i].x);
    for(int i = 1; i <= n; i++)
        b[i].x = (double)1 / i / i,c[i].x = a[n - i].x;
    c[0].x = a[n].x;
    BU(n + n,lim);
    FFT(a,1,lim),FFT(b,1,lim),FFT(c,1,lim);
    for(int i = 0; i <= lim; i++)a[i] = a[i] * b[i];
    for(int i = 0; i <= lim; i++)c[i] = c[i] * b[i];
    FFT(a,-1,lim),FFT(c,-1,lim);
    for(int i = 1; i <= n; i++)
        printf("%.3lf\n",(a[i].x - c[n - i].x) / lim);
    return 0;
}

DNA
巧妙地把每个字母的匹配转化成了函数的卷积，这个公式的推导不像上个题目一样，但是同样也使用了类似的变换

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;


const double Pi=acos(-1.0);
struct cp{
    
    
    double x,y;
    cp (double xx=0,double yy=0){
    
    x=xx,y=yy;}
}a[maxn],b[maxn],c[maxn];
cp operator + (cp a,cp b){
    
     return cp(a.x+b.x , a.y+b.y);}
cp operator - (cp a,cp b){
    
     return cp(a.x-b.x , a.y-b.y);}
cp operator * (cp a,cp b){
    
     return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}

int n,m,r[maxn],t,cnt[maxn];
char str1[maxn],str2[maxn],p[6] = "AGCT";

void FFT(cp *A,int type, int n){
    
    
    for(int i = 0;i < n; i++)
        if(i < r[i]) swap(A[i], A[r[i]]);
    for(int mid = 1;mid < n; mid <<= 1){
    
    
        cp Wn(cos(Pi / mid) , type * sin(Pi / mid));
        for(int R = mid << 1,j = 0;j < n;j += R){
    
    
            cp w(1,0);
            for(int k = 0;k < mid;k++,w = w*Wn){
    
    
                 cp x = A[j + k],y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + mid + k]=x - y;
            }
        }
    }
}

void BU(int sum, int &n){
    
    
    int l = 0;n = 1;
    while(n <= sum)l++, n <<= 1;
    for(int i=0;i<n;i++)
        r[i]= ( r[i>>1]>>1 )| ( (i&1)<<(l-1) ) ;
}
int main(){
    
    

    scanf("%d",&t);
    while(t--){
    
    
        int lim,len1,len2,ans = 0;
        scanf("%s%s",str1,str2);
        len1 = strlen(str1),len2 = strlen(str2);
        BU(len1 + len2, lim);
        for(int i = 0; i <= lim; i++)cnt[i] = 0;
        for(int i = 0; i < 4; i++){
    
    
            for(int j = 0; j < len1; j++)
                a[j].x = (str1[j] == p[i]),a[j].y = 0;
            for(int j = 0; j < len2; j++)
                b[len2 - 1 - j].x = (str2[j] == p[i]),b[len2 - 1 - j].y = 0;
            for(int j = len1; j <= lim; j++)
                a[j].x = a[j].y = 0;
            for(int j = len2; j <= lim; j++)
                b[j].x = b[j].y = 0;

            FFT(a,1,lim),FFT(b,1,lim);
            for(int j = 0; j <= lim; j++)
                a[j] = a[j] * b[j];
            FFT(a,-1,lim);
            for(int j = len2 - 1; j <= len1 - 1; j++)
                cnt[j - len2 + 1] += (int)(a[j].x / lim + 0.5);
        }
        for(int i = 0; i <= len1 - len2; i++)
            if(cnt[i] + 3 >= len2)ans++;
        printf("%d\n",ans);
    }

    return 0;
}

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