hihoCoder 1430 A Boring Problem(数论)

来源：互联网发布：数据对比分析的方法编辑：程序博客网时间：2024/06/16 00:44

https://hihocoder.com/problemset/problem/1430?sid=1202564

分析：

AC代码：

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define gcd(a,b) __gcd(a,b)#define mset(a,n) memset(a,n,sizeof(a))#define FINfreopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;char str[50005];LL sum[50005];LL sum1[50005][105];LL sum2[50005][105];LL C[105][105];void init(){    C[0][0]=1;    for(int i=1;i<=100;i++)        for(int j=0;j<=i;j++)        C[i][j]=(j==0)?1:(C[i-1][j]+C[i-1][j-1])%mod;}int main(){    //FIN;    init();int t;scanf ("%d",&t);while (t--){        LL n,k;        scanf ("%lld%lld",&n,&k);        scanf ("%s",str);        sum[0]=0;        for (int i=1;i<=n;i++)  sum[i]=(sum[i-1]+(LL)(str[i-1]-'0'))%mod;        //        sum1[i][j]表示sum[i]的j次方         for (int i=0;i<=n;i++){            sum1[i][0]=1;            for (int j=1;j<=k;j++){                sum1[i][j]=(sum1[i][j-1]*sum[i])%mod;            }        }        //       sum2[i][j]表示j次方下-sum1的前i项和           for (int i=0;i<=k;i++){            sum2[0][i]=sum1[0][i];            for (int j=1;j<=n;j++){                sum2[j][i]=(sum2[j-1][i]+((i%2==0)?1:-1)*sum1[j][i])%mod;            }        }        for (int i=1;i<=n;i++){            LL ans=0;            for (int j=0;j<=k;j++){                ans=(ans+C[k][j]*sum1[i][j]%mod*sum2[i-1][k-j]%mod+mod)%mod;            }            printf("%lld%c",ans,i==n?'\n':' ');        }}return 0;}

阅读全文

0 0