CodeForces 223C Partial Sums 多次前缀和

Partial Sums

题解：

一个数列多次前缀和之后， 对于第i个数来说他的答案就是

for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= i; ++j){
                b[i] = (b[i] + 1ll * a[j] * C(k-1+j-i,j-i)) % mod;
            }
        }

唯一注意的就是这个k会到1e9。

观察可能，其实我们最多也就用了n个组合数， 并且这个C(n, m) 的 m 足够小。

所以我们可以根据定义先把这几个组合数先预处理出来。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 2e3 + 100;
int n, k;
int a[N], b[N];
int inv[N];
int c[N];
void init(){
    int MOD = mod;
    inv[1] = 1;
    for(int i = 2; i < N; i ++){
        inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;
    }
    c[0] = 1;
    for(int i = 1; i <= n; ++i){
        c[i] = 1;
        int now = k + i - 1;
        for(int j = 1; j <= i; ++j){
            c[i] = 1ll * c[i] * now % mod * inv[j] % mod;
            --now;
        }
    }
}
int main(){
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    init();
    if(k){
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= i; ++j){
                b[i] = (b[i] + 1ll * a[j] * c[i-j]) % mod;
            }
        }
        for(int i = 1; i <= n; ++i)
            a[i] = b[i];
    }
    for(int i = 1; i <= n; ++i){
        printf("%d%c", a[i], " 
"[i==n]);
    }
    return 0;
}