BC18hdoj5104&&hdoj5105&&hdoj5106

Primes Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2014    Accepted Submission(s): 926

Problem Description

Given a number n, please count how many tuple(p1, p2, p3) satisfied that p1<=p2<=p3, p1,p2,p3 are primes and p1 + p2 + p3 = n.

 

Input

Multiple test cases(less than 100), for each test case, the only line indicates the positive integer n(n≤10000).

 

Output

For each test case, print the number of ways.

 

Sample Input

39

 

Sample Output

02

 

Source

BestCoder Round #18

 

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>using namespace std;const int maxn=10010;int num[maxn];int prime[maxn];bool isprime[maxn];int main(){    int n,i,j,k,ans,cnt=0;    memset(isprime,false,sizeof(isprime));    isprime[0]=isprime[1]=true;    for(i=2;i<maxn;++i){        if(isprime[i])continue;        prime[cnt++]=i;        for(j=i*i;j<maxn;j+=i){            isprime[j]=true;        }    }    while(scanf("%d",&n)!=EOF){        int ans=0;//2 3 5 7         for(i=0;i<cnt&&prime[i]<=n;++i){            for(j=i;j<cnt&&prime[i]+prime[j]<=n;++j){                if(!isprime[n-prime[i]-prime[j]]&&(n-prime[i]-prime[j])>=prime[j])ans++;            }        }        printf("%d\n",ans);    }    return 0;}

Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2418    Accepted Submission(s): 579

Problem Description

Here has an function:
  f(x)=|a∗x3+b∗x2+c∗x+d|(L≤x≤R)
Please figure out the maximum result of f(x).

 

Input

Multiple test cases(less than 100). For each test case, there will be only 1 line contains 6 numbers a, b, c, d, L and R. (−10≤a,b,c,d≤10,−100≤L≤R≤100)

 

Output

For each test case, print the answer that was rounded to 2 digits after decimal point in 1 line.

 

Sample Input

1.00 2.00 3.00 4.00 5.00 6.00

 

Sample Output

310.00

 

Source

BestCoder Round #18

 

注意a和b等于0的情况还有精度

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#define eps 1e-10using namespace std;int main(){    double a,b,c,d,l,r;    while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&l,&r)!=EOF){        if(fabs(a)<=eps){            if(fabs(b)<=eps){                printf("%.2lf\n",max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)));            }            else {                double x=-c/(2.0*b);                if(x-l>=-eps&&x-r<=eps){                    printf("%.2lf\n",max(max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)),fabs(a*x*x*x+b*x*x+c*x+d)));                }                else {                    printf("%.2lf\n",max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)));                }            }            continue;        }         double t=4.0*b*b-12.0*a*c;        if(t<=eps){            printf("%.2lf\n",max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)));        }        else {            t=sqrt(t);            double x1=(-2.0*b+t)/(6.0*a);            double x2=(-2.0*b-t)/(6.0*a);            if(x1-l>=-eps&&x1-r<=eps&&x2-l>=-eps&&x2-r<=eps)                printf("%.2lf\n",max(max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)),max(fabs(a*x1*x1*x1+b*x1*x1+c*x1+d),fabs(a*x2*x2*x2+b*x2*x2+c*x2+d))));            else if(x1-l>=-eps&&x1-r<=eps){                printf("%.2lf\n",max(max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)),fabs(a*x1*x1*x1+b*x1*x1+c*x1+d)));            }            else if(x2-l>=-eps&&x2-r<=eps){                printf("%.2lf\n",max(max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)),fabs(a*x2*x2*x2+b*x2*x2+c*x2+d)));            }            else {                printf("%.2lf\n",max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d)));            }        }    }    return 0;}

Bits Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 701    Accepted Submission(s): 183

Problem 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(n≤1000,0<R<21000), 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

7

 

Source

BestCoder Round #18

 

对于一个n-bit数，可以根据与R最高不同位的位置分成几类。比如R=100100010，可以分成0xxxxxxxx,1000xxxxx,10010000x三类。x处可任取0或者1。对于一类数，设与R不同的最高位为L（从低位数起，base 0)，设n0 = n - 高于L位的1的个数。此时和分两部分，高于L位的有CLn0C^{n_0}_LC​L​n​0​​​​乘以L位之前的数字，低于L位的部分有CLn0∗n0/L∗(2L−1)C^{n_0}_L * n0 / L * (2^{L-1})C​L​n​0​​​​∗n0/L∗(2​L−1​​) (因为低于L位的部分每一个位的1的个数相同，1的个数总数有CLn0∗n0C^{n_0}_L*n_0C​L​n​0​​​​∗n​0​​，平均到每个位则是CLn0∗n0/LC^{n_0}_L*n_0/LC​L​n​0​​​​∗n​0​​/L。所以低位部分和有CLn0∗n0/L∗(1+2+...+2L−1)=CLn0∗n0/L∗(2L−1)C^{n_0}_L*n_0/L*(1+2+...+2^{L-1}) = C^{n_0}_L*n_0/L*(2^{L-1})C​L​n​0​​​​∗n​0​​/L∗(1+2+...+2​L−1​​)=C​L​n​0​​​​∗n​0​​/L∗(2​L​​-1) 

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#define MOD 1000000007using namespace std;const int maxn=1010;char bit[maxn];long long Pow[maxn];int C[maxn][maxn];void init(){      Pow[0]=1;      for(int i=0;i<maxn;++i){          C[i][i]=C[i][0]=1;          if(i)Pow[i]=(Pow[i-1]*2)%MOD;    }      for(int i=2;i<maxn;++i){            for(int j=1;j<i;++j){                C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;            }        }  }int main(){    init();    int n,i,j,k;    while(scanf("%d %s",&n,bit)!=EOF){        int len=strlen(bit);        long long pre=0,cnt=0,ans=0;        for(i=0;i<len;++i){            if(bit[i]=='1'){                if((n-cnt)<=(len-i-1)){                    if(n-cnt)ans=(ans+C[len-i-2][n-cnt-1]%MOD*(Pow[len-i-1]-1))%MOD;                    ans=(ans+pre*C[len-i-1][n-cnt])%MOD;                    pre=(pre+Pow[len-i-1])%MOD;cnt++;                }            }            if(cnt>n)break;         }        printf("%lld\n",ans);    }    return 0;}