[acm/icpc2016北京赛区][hihocoder1430] A Boring Problem 推公式

来源：互联网发布：万能数据恢复破解版编辑：程序博客网时间：2024/05/21 06:37

无脑推公式题
一开始推出来O(nk2)的超时了，后来把复杂度降低到O(nk)才跑过去

对于每个答案ans[i]，存在表达式
ans[i]=∑ij=1F[j][i]
令s[i]为前缀和
F[j][i]=(s[i]−s[j−1])k
二项式展开：
F[j][i]=C0ks[i]k+C1ks[i]k−1(−s[j−1])+...+Ckk(−s[j−1])k
把所有Cpks[i]k−p提出来，剩下的用前缀和维护就行了

#include <iostream>#include <cstdio>#define N 100050#define K 105#define mod 1000000007LLusing namespace std;typedef long long LL;int n,k;LL C[K][K],a[N],ans[N],sum[N],S[N][K],T[N][K],r[N][K];char s[N];void inc(LL &x,LL y) { x = (x+y) % mod; }void solve() {    scanf("%d%d",&n,&k);    for (int i=0;i<=k;i++) C[i][0] = C[i][i] = 1LL;    for (int i=1;i<=k;i++)        for (int j=1;j<=i;j++) C[i][j] = ( 0LL + C[i-1][j-1] + C[i-1][j] ) % mod;    scanf("%s",s+1);    for (int i=1;i<=n;i++) a[i] = s[i] - '0';     for (int i=1;i<=n;i++) sum[i] = (sum[i-1] + a[i]) % mod;    // S[i][j]  -->  sum[i] ^ j    for (int i=0;i<=n;i++) S[i][0] = 1LL;    for (int i=1;i<=n;i++)         for (int j=1;j<=k;j++) S[i][j] = S[i][j-1] * sum[i] % mod;    /*    // T[i][j]  -->  (-sum[i]) ^ j    for (int i=0;i<=n;i++) T[i][0] = 1LL;    for (int i=1;i<=n;i++)         for (int j=1;j<=k;j++) {            T[i][j] = T[i][j-1] * (-sum[i]) % mod;            if (T[i][j] < 0) T[i][j] += mod;        }    */    // r[i][j] --> sigma(p=1~i) s[p][j]    for (int j=0;j<=k;j++) r[0][j] = S[0][j];    for (int i=1;i<=n;i++)         for (int j=0;j<=k;j++)             r[i][j] = ( r[i-1][j] + S[i][j] ) % mod;    for (int i=1;i<=n;i++) {        ans[i] = 0LL;        for (int p=0;p<=k;p++)            if (p%2 == 0)                inc( ans[i] ,  C[k][p] * S[i][k-p] % mod * r[i-1][p] % mod );            else                inc( ans[i] , -C[k][p] * S[i][k-p] % mod * r[i-1][p] % mod );         if (ans[i] < 0) ans[i] += mod;    }    for (int i=1;i<=n;i++) printf("%d%c",(int)ans[i],i==n?'\n':' ');    return ;}int main() {    int T = 0; scanf("%d",&T);    for (int _=1;_<=T;_++) solve();    return 0;}

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