题解 表格
推出来柿子了，但不会展开，就挂了
- 一个 \(n\) 次多项式的前缀和是 \(n+1\) 次多项式
- 当发现一个柿子不会展开时，可以先大致确定它是几次多项式，然后直接高斯消元出系数
然后这题就没了
正解好像是拉格朗日插值，先咕了
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
ll n, m, l, r;
const ll mod=1e9+7;
namespace force{
int ans;
void solve() {
for (int x1=1; x1<=n; ++x1)
for (int x2=x1+1; x2<=n; ++x2) if (x2!=x1)
for (int x3=x2+1; x3<=n; ++x3) if (x3!=x1 && x3!=x2)
for (int y1=1; y1<=m; ++y1)
for (int y2=1; y2<=m; ++y2) if (y2!=y1)
for (int y3=1; y3<=m; ++y3) if (y3!=y2 && y3!=y1) {
int dis=abs(x1-x2)+abs(y1-y2)+abs(x2-x3)+abs(y2-y3)+abs(x1-x3)+abs(y1-y3);
if (l<=dis && dis<=r) {
++ans;
// cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<' '<<x3<<' '<<y3<<endl;
}
}
printf("%lld\n", ans%mod);
exit(0);
}
}
namespace task1{
ll fac[N], inv[N], ans;
inline ll C(int n, int k) {return fac[n]*inv[k]%mod*inv[n-k]%mod;}
void solve() {
int lim=max(n, m);
fac[0]=fac[1]=1; inv[0]=inv[1]=1;
for (int i=2; i<=lim; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) inv[i]=inv[i-1]*inv[i]%mod;
for (int i=3; i<=n; ++i)
for (int j=3; j<=m; ++j)
if (l<=(i+j-2)*2 && (i+j-2)*2<=r)
ans=(ans+(n-i+1ll)*(m-j+1ll)%mod*(i-2ll)%mod*(j-2ll)%mod*fac[3]%mod)%mod;
printf("%lld\n", ans);
exit(0);
}
}
namespace task2{
ll fac[N], inv[N], inv2[N];
ll ans;
// inline ll C(int n, int k) {return fac[n]*inv2[k]%mod*inv2[n-k]%mod;}
inline ll sum2(ll k) {return k*(k+1)%mod*(2ll*k+1)%mod*inv[6]%mod;}
inline ll sum(ll k) {return k*(k+1)%mod*inv[2]%mod;}
void solve() {
// int lim=max(max(n, m), 10ll);
int lim=10;
fac[0]=fac[1]=1; inv[0]=inv[1]=1; inv2[0]=inv2[1]=1;
for (int i=2; i<=lim; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) inv2[i]=inv2[i-1]*inv[i]%mod;
for (int i=3; i<=n; ++i) {
ll L=max(3ll, (l-2*i+5ll)/2ll), R=min(m, (r-2ll*i+4ll)/2ll);
if (L<=R) ans = (ans+1ll*(n-i+1ll)*(i-2ll)%mod * (( -(sum2(R)-sum2(L-1)) + (sum(R)-sum(L-1))%mod*(m+3)%mod - (2ll*m+2)*(R-L+1ll)%mod )%mod)%mod)%mod;
}
// for (int i=3; i<=n; ++i) {
// cout<<"i: "<<i<<' '<<max(3, (l-2*i+5)/2)<<' '<<min(m, (r-2*i+4)/2)<<endl;
// }
ans = ans*fac[3]%mod;
printf("%lld\n", (ans%mod+mod)%mod);
exit(0);
}
}
namespace task3{
ll fac[N], inv[N], ans;
inline ll C(int n, int k) {return fac[n]*inv[k]%mod*inv[n-k]%mod;}
void solve() {
int lim=max(max(n, m), 10ll);
fac[0]=fac[1]=1; inv[0]=inv[1]=1;
for (int i=2; i<=lim; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) inv[i]=inv[i-1]*inv[i]%mod;
for (int i=3; i<=n; ++i)
for (int j=max(3ll, (l-2*i+5)/2),ends=min(m, (r-2*i+4)/2); j<=ends; ++j)
ans=(ans+(n-i+1ll)*(m-j+1ll)%mod*(i-2ll)%mod*(j-2ll)%mod*fac[3]%mod)%mod;
// for (int i=3; i<=n; ++i) {
// cout<<"i: "<<i<<' '<<max(3, (l-2*i+5)/2)<<' '<<min(m, (r-2*i+4)/2)<<endl;
// }
printf("%lld\n", ans);
exit(0);
}
}
namespace task{
ll fac[N], inv[N], inv2[N], k[10], Gm, r;
inline ll sum2(ll k) {k%=mod; return k*(k+1)%mod*(2ll*k%mod+1)%mod*inv[6]%mod;}
inline ll sum(ll k) {k%=mod; return k*(k+1)%mod*inv[2]%mod;}
inline ll G(ll k) {return (-(sum2(k)-5ll)+(sum(k)-3ll)%mod*(m%mod+3ll)%mod-(2ll*m%mod+2ll)*(k%mod-2ll)%mod)%mod;}
inline ll H1(ll base) {base%=mod; ll ans=0ll,t=1ll; for (int i=1; i<=7; ++i,t=t*base%mod) ans=(ans+k[i]*t%mod)%mod; return ans;}
inline ll H2(ll k) {return (-(sum2(k)-5ll)+(sum(k)-3ll)%mod*(n%mod+3ll)%mod-(2ll*n%mod+2ll)*(k%mod-2ll)%mod)%mod*Gm%mod;}
inline ll qpow(ll a, ll b) {a%=mod; ll ans=1ll; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
struct matrix{
int n, m;
ll a[10][10];
inline void resize(int s, int t) {n=s; m=t; memset(a, 0, sizeof(a));}
inline void put() {for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) cout<<a[i][j]<<' '; cout<<endl;}}
inline ll* operator [] (int t) {return a[t];}
void gauss(ll ans[]) {
for (int i=1; i<=n; ++i) {
int r=i;
for (int j=i+1; j<=n; ++j) if (llabs(a[j][i])>llabs(a[r][i])) r=j;
swap(a[i], a[r]);
for (int j=1; j<=n; ++j) {
if (i==j) continue;
ll d=a[j][i]*qpow(a[i][i], mod-2)%mod;
for (int k=i; k<=m; ++k) {
a[j][k]=(a[j][k]-a[i][k]*d%mod)%mod;
a[j][k]=(a[j][k]%mod+mod)%mod;
}
}
}
for (int i=1; i<=n; ++i) ans[i]=a[i][m]*qpow(a[i][i], mod-2)%mod;
}
}mat;
ll solve(ll now) {
r=now;
int lim=10;
fac[0]=fac[1]=1; inv[0]=inv[1]=1; inv2[0]=inv2[1]=1;
for (int i=2; i<=lim; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) inv2[i]=inv2[i-1]*inv[i]%mod;
Gm=G(m); mat.resize(7, 8);
// cout<<"G: "<<Gm<<endl;
for (int i=3; i<=9; ++i) {
ll tem=1ll;
for (int j=1; j<=7; ++j,tem=tem*i%mod) mat[i-2][j]=tem;
tem=0;
for (int j=3; j<=i; ++j) tem=((tem+1ll*(j-2)*(n%mod-j+1)%mod*G(r/2-j+2)%mod)%mod+mod)%mod;
mat[i-2][8]=tem;
}
#if 0
mat.put(); cout<<endl;
mat.gauss(k);
mat.put(); cout<<endl;
for (int i=3; i<=15; ++i) {
ll tem=0;
for (int j=3; j<=i; ++j) tem=((tem+1ll*(j-2)*(n-j+1)%mod*G(r/2-j+2)%mod)%mod+mod)%mod;
cout<<"H1: "<<H1(i)<<' '<<tem<<endl;
}
for (int i=3; i<=15; ++i) {
ll tem=0;
for (int j=3; j<=i; ++j) tem=(tem+1ll*(j-2)*(m-j+1)%mod)%mod;
// cout<<"G: "<<G(i)<<' '<<tem<<endl;
}
for (int i=3; i<=15; ++i) {
ll tem=0;
for (int j=3; j<=i; ++j) tem=(tem+1ll*(j-2)*(n-j+1)%mod*G(m)%mod)%mod;
cout<<"H2: "<<H2(i)<<' '<<tem<<endl;
}
#else
mat.gauss(k);
#endif
// cout<<H1(10000)<<endl;
ll t=r/2-m+2;
ll x=min(n, r/2-1);
ll ans=0;
// cout<<"xt: "<<x<<' '<<t<<' '<<(H1(min(x, t))%mod+mod)%mod<<endl;
if (x>=t) {
if (t>3) ans = (ans+H1(x)-H1(t-1))%mod;
else ans = (ans+H1(x))%mod;
if (t>3) ans = (ans+H2(t-1))%mod;
return ans;
}
else {
if (x>=3) return H2(x)%mod;
// else return H2(3)%mod;
else return 0;
}
// cout<<t<<' '<<x<<endl;
// cout<<H1(t)-H1(t-1)<<' '<<H2(t)-H2(t-1)<<endl;
// cout<<H1(x)<<' '<<H1(t-1)<<' '<<H2(t-1)<<endl;
// return (H1(x)-H1(t-1)+H2(t-1))%mod;
}
}
signed main()
{
freopen("table.in", "r", stdin);
freopen("table.out", "w", stdout);
n=read(); m=read(); l=read(); r=read();
// force::solve();
// cout<<task::solve(r)<<endl;
// cout<<task::solve(l-1)<<endl;
printf("%lld\n", ((task::solve(r)-task::solve(l-1))%mod*6ll%mod+mod)%mod);
return 0;
}
