HDU 5106 Bits Problem（数位DP->二进制大数模拟）@

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Bits Problem

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

If the quantity of '1' in a number's binary digits is n, we call this number a n-onebit number. For instance, 8(1000) is a 1-onebit number, and 5(101) is a 2-onebit number. Now give you a number - n, please figure out the sum of n-onebit number belong to [0, R).

Input

Multiple test cases(less than 65). For each test case, there will only 1 line contains a non-negative integer n and a positive integer

, R is represented by binary digits, the data guarantee that there is no leading zeros.

Output

For each test case, print the answer module 1000000007 in one line.

Sample Input

1 1000

Sample Output

一开始想用三维数组来表示，但是情况很难考虑清楚就开了俩个二维数组一个表示数量，一个表示和，需要预处理，因为不包括右端点，中间的处理过程还需要一定的技巧。

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<cmath>using namespace std;typedef long long LL;const int mod = 1000000007;const int N = 1005;LL dp[N][N], ct[N][N], b[N];char str[N];int num[N];int main(){    b[0]=1;    for(int i=1;i<N;i++)        b[i]=(b[i-1]*2)%mod;    ct[0][0]=1;    for(int i=1;i<N;i++)    {        ct[i][0]=1;        for(int j=1;j<=i;j++)        {            ct[i][j]=(ct[i-1][j-1]+ct[i-1][j])%mod;        }    }    memset(dp,0,sizeof(dp));    for(int i=1;i<N;i++)    {        for(int j=1;j<=i;j++)        {            dp[i][j]=(dp[i-1][j-1]+dp[i-1][j]+ct[i-1][j-1]*b[i-1]%mod)%mod;        }    }    int n;    while(scanf("%d %s",&n, str)!=EOF)    {        int len = strlen(str);        memset(num,0,sizeof(num));        for(int i=0;i<len;i++)        {            num[i+1]=(str[len-i-1]-'0');        }        LL ans=0, tmp=0;        int cnt=0;        for(int i=len;i>0;i--)        {            if(cnt>n)                break;            if(num[i])            {                ans+=((ct[i-1][n-cnt]*tmp%mod+dp[i-1][n-cnt])%mod);                ans%=mod;                tmp+=b[i-1];                tmp%=mod;                cnt++;            }        }        printf("%I64d\n",ans);    }    return 0;}

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