拉格朗日插值

study from

https://blog.csdn.net/Code92007/article/details/94412729

P5050 【模板】多项式多点求值

各种神仙常数优化、NTT

1.常数优化(来自洛谷题解)

学习了！

O2优化 g++某些版本

循环展开：并行操作

http://blog.miskcoo.com/2015/05/polynomial-multipoint-eval-and-interpolation

但还是超时了

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ull unsigned long long
#define R register

const double eps=1e-8;
const ull inf=1e9;
const ull mod=998244353;
const int maxn=1e5+10;

ull a[maxn],b[17]; ///can not use as constant

inline void read(R ull &y)
{
    char ch=getchar();
    y=0;
    while (ch<'0' || ch>'9')
        ch=getchar();
    while (ch>='0' && ch<='9')
        y=y*10+ch-48,ch=getchar();
}

char pr[10];
int cnt;
inline void write(R ull &y)
{
    cnt=0;
    while (y)
    {
        pr[++cnt]=y%10+48;
        y/=10;
    }
    for (R int i=cnt;i;i--)
        putchar(pr[i]);
    putchar('
');
}

int main()
{
    R ull n,m,x,y,c1,c2,c3,c4;
    R int i;
    read(n),read(m);
    for (i=0;i<=n;i++)
        read(a[i]);
    while (m--)
    {
        read(x);
        b[0]=1;
        for (i=1;i<=16;i++)
            b[i]=b[i-1]*x%mod;
        y=0;
        for (i=n;i>=15;i-=16)
        {
            ///998244353^2*16 ‭15,943,868,612,742,217,744‬
            ///unsigned long long 1.8e19
            c1=a[i]*b[15]+a[i-1]*b[14]+a[i-2]*b[13]+a[i-3]*b[12];
            c2=a[i-4]*b[11]+a[i-5]*b[10]+a[i-6]*b[9]+a[i-7]*b[8];
            c3=a[i-8]*b[7]+a[i-9]*b[6]+a[i-10]*b[5]+a[i-11]*b[4];
            c4=a[i-12]*b[3]+a[i-13]*b[2]+a[i-14]*b[1]+a[i-15];
            y=(c1+c2+c3+c4+y*b[16])%mod;
        }
        for (;i>=0;i--)
            y=(y*x+a[i])%mod;
        write(y);
    }
    return 0;
}
/*
20 3
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 1 3
*/

 原始代码

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long

const double eps=1e-8;
const ll inf=1e9;
const ll mod=998244353;
const int maxn=1e5+10;

ll a[maxn];

inline void read(ll &y)
{
    char ch=getchar();
    y=0;
    while (ch<'0' || ch>'9')
        ch=getchar();
    while (ch>='0' && ch<='9')
        y=y*10+ch-48,ch=getchar();
}

int main()
{
    int n,m,i;
    ll x,y,z;
    scanf("%d%d",&n,&m);
    for (i=0;i<=n;i++)
        read(a[i]);
    while (m--)
    {
        read(x);
        y=0;
        z=1;
        for (i=0;i<=n;i++)
        {
            (y+=a[i]*z)%=mod;
            z=z*x%mod;
        }
        printf("%lld
",y);
    }
    return 0;
}

P4781 【模板】拉格朗日插值

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long

const double eps=1e-8;
const ll inf=1e9;
const ll mod=998244353;
const int maxn=2e3+10;

ll a[maxn],x[maxn],y[maxn];

ll mul(ll a,ll b)
{
    ll y=1;
    while (b)
    {
        if (b&1)
            y=y*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return y;
}

int main()
{
    ll i,j,n,d,tot,sum=0,u,v;
    scanf("%lld%lld",&n,&d);
    for (i=0;i<n;i++)
        scanf("%lld%lld",&x[i],&y[i]);
    ///n*n
    ///multiply n-1 numbers
    for (i=0;i<n;i++)
    {
        ///can also multiply n numbers and divide one number
        u=1;
        for (j=0;j<n;j++)
            if (i!=j)
                u=u*(d-x[j])%mod;

        v=1;
        for (j=0;j<n;j++)
            if (i!=j)
                v=v*(x[i]-x[j])%mod;

        tot=y[i]*u%mod;
        if (v<0)
            tot=-tot,v=-v;

        (sum+=tot*mul(v,mod-2)%mod)%=mod;
    }
    printf("%lld
",(sum+mod)%mod);
    return 0;
}
/*
3 4
1 4
2 9
3 16
*/

https://loj.ac/problem/166

P5158 【模板】多项式快速插值

之后再补

===============================

自然数幂和

https://codeforces.com/problemset/problem/622/F

F. The Sum of the k-th Powers

k+1次函数的证明

1.

“差分” 牛顿插值多项式

(from https://www.bbsmax.com/A/RnJWLeoE5q/)

2.

不同方法 https://blog.csdn.net/werkeytom_ftd/article/details/50741165

其中的分治fft棒棒的！

伯努利数 https://blog.csdn.net/cj1064789374/article/details/85388995

斯特林数 https://blog.csdn.net/lyd_7_29/article/details/75041818

若干个k次函数相加为k+1次函数

ai

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long

const double eps=1e-8;
const ll inf=1e9;
const ll mod=1e9+7;
const int maxn=1e6+10;

ll y[maxn],a[maxn],jie[maxn],inv_jie[maxn];

ll mul(ll a,ll b)
{
    ll y=1;
    while (b)
    {
        if (b&1)
            y=y*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return y;
}

int main()
{
    ll n,k,i,u,tot,sum=0;
    scanf("%lld%lld",&n,&k);
    k++;
    for (i=1;i<=k;i++)
        y[i]=(y[i-1]+mul(i,k-1))%mod;   ///
    if (n<=k)
    {
        printf("%lld",y[n]);
        return 0;
    }

    u=1;
    for (i=n;i>=n-k;i--)
        u=u*i%mod;

    jie[0]=1;
    for (i=1;i<=k;i++)
        jie[i]=jie[i-1]*i%mod;
    inv_jie[k]=mul(jie[k],mod-2);
    for (i=k-1;i>=0;i--)    ///
        inv_jie[i]=inv_jie[i+1]*(i+1)%mod;

    for (i=0;i<=k;i++)
    {
        tot=inv_jie[i]*inv_jie[k-i]%mod *u%mod *mul(n-i,mod-2)%mod;
        if ((k-i)&1)
            tot=-tot;
        (sum+=tot*y[i])%=mod;
    }
    printf("%lld",(sum+mod)%mod);
    return 0;
}
/*
1 10
10 1
10 2
10 3
*/

另外的题目

The 2019 ICPC China Nanchang National Invitational and International Silk-Road Programming Contest

https://www.cnblogs.com/cmyg/p/11255754.html

BZOJ2665
BZOJ4559