BC#45 1004 && HDU 5275【差值方法】

偏执的太偏执、 2022-08-02 12:53 142阅读 0赞

作为BC的1004，比赛的时候确实没什么人做出来，确实用差值的方法不太容易想到  
[差值方法介绍][Link 1]  
用拉格朗日差值来做

// whn6325689
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    #define root 1,1,n
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    const int MOD=1000000007;
    const int MAXN=3010;
    
    int p[MAXN][2],mul[MAXN][MAXN];
    int inv[250010];
    int n,m;
    
    ll power(int a,int n)
    {
        ll ans=1,tmp=a;
        while(n)
        {
            if(n&1)
                ans=ans*tmp%MOD;
            tmp=tmp*tmp%MOD;
            n>>=1;
        }
        return ans;
    }
    
    void preprocess()
    {
        for(int i=1;i<=n;i++)
        {
            mul[i][i]=1;
            for(int j=i+1;j<=n;j++)
            {
                mul[i][j]=1LL*mul[i][j-1]*inv[abs(p[i][0]-p[j][0])]%MOD*(p[i][0]<p[j][0]?-1:1);
                if(mul[i][j]<0) mul[i][j]+=MOD;
            }
            for(int j=i-1;j>=1;j--)
            {
                mul[i][j]=1LL*mul[i][j+1]*inv[abs(p[i][0]-p[j][0])]%MOD*(p[i][0]<p[j][0]?-1:1);
                if(mul[i][j]<0) mul[i][j]+=MOD;
            }
        }
    }
    
    int main()
    {
        int l,r,x;
        inv[0]=1;
        for(int i=1;i<250005;i++)
            inv[i]=power(i,MOD-2);
        while(read(n))
        {
            for(int i=1;i<=n;i++)
                read(p[i][0]),read(p[i][1]);
            preprocess();
            read(m);
            while(m--)
            {
                read(l),read(r),read(x);
                int res=0,tot=1;
                for(int i=l;i<=r;i++)
                    tot=1LL*tot*(x-p[i][0])%MOD;
                for(int i=l;i<=r;i++)
                {
                    int tmp=1LL*p[i][1]*mul[i][l]%MOD*mul[i][r]%MOD;
                    if(x!=p[i][0])
                        tmp=1LL*tmp*tot%MOD*inv[abs(x-p[i][0])]%MOD*(x<p[i][0]?-1:1);
                    else
                    {
                        int t=1;
                        for(int j=l;j<=r;j++)
                            if(i!=j)
                                t=1LL*t*(x-p[j][0])%MOD;
                        tmp=1LL*tmp*t%MOD;
                    }
                    if(tmp<0)   tmp+=MOD;
                    //cout<<tmp<<endl;
                    res=(res+tmp)%MOD;
                }
                printf("%d\n",res);
            }
        }
        return 0;
    }

用牛顿差值来做

#include<cstdio>
    #define N 3010
    #define p 1000000007
    int n,q,x[N],f[N][N],inv[250010];
    #define getinv(x) (x>=0?inv[x]:p-inv[-(x)])
    int main(){
        scanf("%d",&n);inv[1]=1;
        for(int i=2;i<=250000;i++)
            inv[i]=1LL*(p-p/i)*inv[p%i]%p;
        for(int i=1;i<=n;i++)
            scanf("%d%d",x+i,f[i]+1);
        for(int i=2;i<=n;i++)
            for(int j=2,k=i-1;j<=i;j++,k--)
                f[i][j]=1LL*(p+f[i-1][j-1]-f[i][j-1])*getinv(x[k]-x[i])%p;
        scanf("%d",&q);
        for(int l,r,x0,ans;q--;printf("%d\n",ans)){
            scanf("%d%d%d",&l,&r,&x0);ans=0;
            for(int i=l,j=1,sum=1;i<=r;i++,j++)
                ans=(ans+1LL*f[i][j]*sum)%p,sum=1LL*sum*(p+x0-x[i])%p;
        }
    }

[Link 1]: http://blog.csdn.net/u013007900/article/details/46559131