题意 :
分析 :
分析就直接参考这个链接吧 ==> Click here
大体的思路就是
求和顺序不影响结果、故转化一下思路枚举每个最大值对答案的贡献最后累加就是结果
期间计数的过程要用到容斥和多项式求和 ( 利用拉格朗日求即可 ) 具体参考给出的链接
#include<bits/stdc++.h>
#define LL long long
#define ULL unsigned long long
#define scl(i) scanf("%lld", &i)
#define scll(i, j) scanf("%lld %lld", &i, &j)
#define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k)
#define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l)
#define scs(i) scanf("%s", i)
#define sci(i) scanf("%d", &i)
#define scd(i) scanf("%lf", &i)
#define scIl(i) scanf("%I64d", &i)
#define scii(i, j) scanf("%d %d", &i, &j)
#define scdd(i, j) scanf("%lf %lf", &i, &j)
#define scIll(i, j) scanf("%I64d %I64d", &i, &j)
#define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k)
#define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k)
#define scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k)
#define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l)
#define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l)
#define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define lowbit(i) (i & (-i))
#define mem(i, j) memset(i, j, sizeof(i))
#define fir first
#define sec second
#define VI vector<int>
#define ins(i) insert(i)
#define pb(i) push_back(i)
#define pii pair<int, int>
#define VL vector<long long>
#define mk(i, j) make_pair(i, j)
#define all(i) i.begin(), i.end()
#define pll pair<long long, long long>
#define _TIME 0
#define _INPUT 0
#define _OUTPUT 0
clock_t START, END;
void __stTIME();
void __enTIME();
void __IOPUT();
using namespace std;
;
;
LL pow_mod(LL a, LL b)
{
a %= mod;
LL ret = ;
while(b){
) ret = (ret * a) % mod;
a = ( a * a ) % mod;
b >>= ;
}
return ret;
}
namespace polysum {
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
;
LL a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][],C[D];
LL powmod(LL a,LL b){LL res=;a%=mod;assert(b>=);){)res=res*a%mod;a=a*a%mod;}return res;}
LL calcn(int d,LL *a,LL n) { // a[0].. a[d] a[n]
if (n<=d) return a[n];
p1[]=p2[]=;
rep(i,,d+) {
LL t=(n-i+mod)%mod;
p1[i+]=p1[i]*t%mod;
}
rep(i,,d+) {
LL t=(n-d+i+mod)%mod;
p2[i+]=p2[i]*t%mod;
}
LL ans=;
rep(i,,d+) {
LL t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
) ans=(ans-t+mod)%mod;
else ans=(ans+t)%mod;
}
return ans;
}
void init(int M) {
f[]=f[]=g[]=g[]=;
rep(i,,M+) f[i]=f[i-]*i%mod;
g[M+]=powmod(f[M+],mod-);
per(i,,M+) g[i]=g[i+]*(i+)%mod;
}
LL polysum(LL m,LL *a,LL n) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
LL b[D];
;i<=m;i++) b[i]=a[i];
b[m+]=calcn(m,b,m+);
rep(i,,m+) b[i]=(b[i-]+b[i])%mod;
,b,n-);
}
LL qpolysum(LL R,LL n,LL *a,LL m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
) return polysum(n,a,m);
a[m+]=calcn(m,a,m+);
LL r=powmod(R,mod-),p3=,p4=,c,ans;
h[][]=;h[][]=;
rep(i,,m+) {
h[i][]=(h[i-][]+a[i-])*r%mod;
h[i][]=h[i-][]*r%mod;
}
rep(i,,m+) {
LL t=g[i]*g[m+-i]%mod;
) p3=((p3-h[i][]*t)%mod+mod)%mod,p4=((p4-h[i][]*t)%mod+mod)%mod;
]*t)%mod,p4=(p4+h[i][]*t)%mod;
}
c=powmod(p4,mod-)*(mod-p3)%mod;
rep(i,,m+) h[i][]=(h[i][]+h[i][]*c)%mod;
rep(i,,m+) C[i]=h[i][];
ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
) ans+=mod;
return ans;
}
}
LL arr[maxn];
int main(void){__stTIME();__IOPUT();
int n;
polysum::init(maxn);
while(~sci(n)){
; i<=n; i++) scl(arr[i]);
sort(arr+, arr++n);
arr[] = ;
LL now = ;
LL b[maxn];
LL ans = ;
; i<=n; i++){
]){
now = (now * arr[i]) % mod;
continue;
}
b[] = ;
; j<=n-i+; j++)
b[j] = (LL)j * ((pow_mod((LL)j, n-i+) - pow_mod((LL)j-1LL, n-i+)%mod)+mod)%mod;
LL tmp = ((polysum::polysum(n-i+, b, arr[i]+) -
polysum::polysum(n-i+, b, arr[i-]+)%mod)+mod)%mod;
ans = (ans + (tmp%mod * now%mod)%mod)%mod;
now = (now * arr[i]) % mod;
}
printf("%lld\n", ans);
}
__enTIME();;}
void __stTIME()
{
#if _TIME
START = clock();
#endif
}
void __enTIME()
{
#if _TIME
END = clock();
cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl;
#endif
}
void __IOPUT()
{
#if _INPUT
freopen("in.txt", "r", stdin);
#endif
#if _OUTPUT
freopen("out.txt", "w", stdout);
#endif
}注 :
N + 1 个点能确定一个 N 次多项式、故拉格朗日插值需要确定 ( 最高次次数 + 1 ) 个点的值
namespace polysum {
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
;//可能需要用到的最高次
LL a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][],C[D];
LL powmod(LL a,LL b){LL res=;a%=mod;assert(b>=);){)res=res*a%mod;a=a*a%mod;}return res;}
LL calcn(int d,LL *a,LL n) { // a[0].. a[d] a[n]
if (n<=d) return a[n];
p1[]=p2[]=;
rep(i,,d+) {
LL t=(n-i+mod)%mod;
p1[i+]=p1[i]*t%mod;
}
rep(i,,d+) {
LL t=(n-d+i+mod)%mod;
p2[i+]=p2[i]*t%mod;
}
LL ans=;
rep(i,,d+) {
LL t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
) ans=(ans-t+mod)%mod;
else ans=(ans+t)%mod;
}
return ans;
}
void init(int M) {//用到的最高次
f[]=f[]=g[]=g[]=;
rep(i,,M+) f[i]=f[i-]*i%mod;
g[M+]=powmod(f[M+],mod-);
per(i,,M+) g[i]=g[i+]*(i+)%mod;
}
LL polysum(LL m,LL *a,LL n) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
;i<=m;i++) b[i]=a[i];
b[m+]=calcn(m,b,m+);
rep(i,,m+) b[i]=(b[i-]+b[i])%mod;
,b,n-);
}
LL qpolysum(LL R,LL n,LL *a,LL m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
) return polysum(n,a,m);
a[m+]=calcn(m,a,m+);
LL r=powmod(R,mod-),p3=,p4=,c,ans;
h[][]=;h[][]=;
rep(i,,m+) {
h[i][]=(h[i-][]+a[i-])*r%mod;
h[i][]=h[i-][]*r%mod;
}
rep(i,,m+) {
LL t=g[i]*g[m+-i]%mod;
) p3=((p3-h[i][]*t)%mod+mod)%mod,p4=((p4-h[i][]*t)%mod+mod)%mod;
]*t)%mod,p4=(p4+h[i][]*t)%mod;
}
c=powmod(p4,mod-)*(mod-p3)%mod;
rep(i,,m+) h[i][]=(h[i][]+h[i][]*c)%mod;
rep(i,,m+) C[i]=h[i][];
ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
) ans+=mod;
return ans;
}
}拉格朗日插值模板 (dls版)
-------------------------------分 割 线-------------------------------
链接题解用到的化简容斥的多项式展开式如下
( x - 1 ) ^ k
= x^k
+ C(k, 1) * x^(k-1) * (-1)^1
+ C(k, 2) * x^(k-2) * (-1)^2
+ ......
+ C(k, k) * x^(k-k) * (-1)^k