Ones and Twos


Problem Statement :


You are using at most A number of 1s and at most B number of 2s. How many different evaluation results are possible when they are formed in an expression containing only addition + sign and multiplication * sign are allowed?

Note that, multiplication takes precedence over addition.

For example, if A=2 and B=2, then we have the following expressions:

1, 1*1 = 1
2, 1*2, 1*1*2, 1+1 = 2
1+2, 1+1*2 = 3
2+2, 2*2, 1+1+2, 1*2*2, 1*1*2*2, 1*2+1*2, 1*1*2+2, 1*2+2 = 4
1+2+2, 1+1*2+2 = 5
1+1+2+2, 1+1+2*2 = 6
So there are 6 unique results that can be formed if A = 2 and B = 2.

Input Format

The first line contains the number of test cases T, T testcases follow each in a newline.
Each testcase contains 2 integers A and B separated by a single space.

Constraints

1 <= T <= 105
0<=A<=1000000000
0<=B<=1000

Output Format

Print the number of different evaluations modulo (%) (109+7.)



Solution :



title-img


                            Solution in C :

In C++ :





#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int MOD = 1e9 + 7;
int dp_sum[1001][1001],dp_bit[1001][1001];
void add(int &x,long long v){
    v%=MOD;
    x+=v;
    if(x>=MOD)x-=MOD;
}
int f_bit(int lv,int B);
int f_sum(int lv,int B){
    if(lv>B)return 1;
    if(dp_sum[lv][B]!=-1)return dp_sum[lv][B];
    int tmp=f_sum(lv+1,B);
    add(tmp,f_bit(lv,B));
    return dp_sum[lv][B]=tmp;

}
int f_bit(int lv,int B){
    if(lv>B)return 0;
    if(dp_bit[lv][B]!=-1)return dp_bit[lv][B];
    return dp_bit[lv][B]=f_sum(lv+1,B-lv);
}
int main(){
    int T,A,B;
    scanf("%d",&T);
    memset(dp_sum,-1,sizeof(dp_sum));
    memset(dp_bit,-1,sizeof(dp_bit));
    while(T--){
        scanf("%d%d",&A,&B);
        int an=0;
        if(A==0)an=f_sum(1,B);
        else{
            int k=1;
            while((1LL<<k)<=A)k++;
            k++;
            add(an,(long long)(A+1)*f_sum(k,B));
            long long ha=(1LL<<k)-A-1;
            int now=0,i=1;
            long long last=0;
            while(ha>0){
                now++;
                if(B<now)break;
                add(an,min((1LL<<i)-last,ha)*f_sum(k,B-now));
                ha-=(1LL<<i)-last;
                last=1LL<<i;
                i++;
                if(i==k){
                    last=0;
                    i=1;
                }
            }
            /*
            long long last=A;
            for(int i=1;i<=B;i++){
                if(last+1>=(1LL<<k))break;
                add(an,(min(((long long)A+(1LL<<i)),(1LL<<k)-1)-last)*f_sum(k,B-i));
                last=min((long long)A+(1LL<<i),(1LL<<k)-1);
            }*/
        }
        printf("%d\n",(an+MOD-1)%MOD);
    }
    return 0;
}








In Java :





import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Solution {

 static Solution main;

 public class Range implements Comparable<Range> {
  int minPower;
  int maxPower;
  long range;
  public Range(int min , int max , long r) {
   minPower = min;
   maxPower = max;
   range = r;
  }
  public int compareTo(Range r) {
   int comp = new Integer(minPower).compareTo(r.minPower);
   if(comp!=0) {
    return comp;
   }
   return  new Integer(maxPower).compareTo(r.maxPower);
  }
 }

 public static void main(String[] args) {
  main = new Solution();
  long [][] dp = new long[1001][1001];
  long [][] dp1 = new long[1001][1001];
  long [][] dpsum = new long[1001][1001];
  long sp = 1000000007;
  for(int i = 0 ; i < 1001 ; i++) {
   Arrays.fill(dp[i], 0);
  }
  long [] pow2 = new long [32];
  pow2[0] = 1;
  for(int i = 1 ; i <32 ; i++) {
   pow2[i ] = pow2[i-1] * 2;
  }
  dp[1][1] = 1;
  for(int i = 2 ; i < 1001 ; i++) {
   for(int j = 1 ; j <= i ; j++) {
    long temp = 0;
    for(int k = j+1 ; k < i ; k++) {
     temp += dp[i-j][k];
     if(temp>=sp) {
      temp-=sp;
     }
    }
    if(i==j) {
     temp++;
     if(temp>=sp) {
      temp-=sp;
     }
    }
    dp[i][j] = temp ;
   }
  }
  for(int k = 0 ; k < 1001 ; k++) {
   Arrays.fill(dp1[k],0);
  }
  for(int i = 0 ; i < 1001 ; i++) {
   dpsum[i][0] = 0;
  }
  for(int i = 0 ; i < 1001 ; i++) {
   for(int j = 1 ; j < 1001 ; j++) {
    dpsum[i][j] = dpsum[i][j-1] + dp[i][j];
    if(dpsum[i][j]>=sp) {
     dpsum[i][j] -= sp;
    }
   }
  }
  for(int k = 1 ; k < 1001 ; k++) {
   for(int i = k ; i < 1001 ; i++) {
    if(i==k) {
     dp1[i][k] = 1;
    } else {
     dp1[i][k] = dp1[i-1][k] + dp[i][k];
     if(dp1[i][k] >= sp) {
      dp1[i][k] -= sp;
     }
    }
   }
  }
  long [][][] list = new long[1001][32][32];
  long [] all = new long[1001];
  for(int i = 0 ; i < 1001 ; i++) {
   for(int j = 0 ; j < 32 ; j++) {
    for(int k = 0 ; k < 32 ; k++) {
     list[i][j][k] = 0;
    }
   }
   all[i] = 0;
   for(int j = 1 ; j <= Math.min(i/2, 500) ; j++) {
    for(int k = j + 1 ; k <= i-j ; k++) {
     long repValue = 0;
     if(j+k==i) {
      repValue ++;
     }
     repValue += dpsum[i-j-k][i-j-k] - dpsum[i-j-k][k];
     if(repValue<0) {
      repValue += sp;
     }
     if(repValue>=sp) {
      repValue -= sp;
     }
     if(k<32) {
      list[i][j][k] = repValue;
     } else {
      all[i] += repValue;
     }
    }
   }
   all[i] %= sp;
  }
  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  BufferedOutputStream bos = new BufferedOutputStream(System.out);
  String eol = System.getProperty("line.separator");
  byte[] eolb = eol.getBytes();
  try {
   String str = br.readLine();
   int t = Integer.parseInt(str);
   for(int i = 0 ; i < t ; i++) {
    str = br.readLine();
    int blank = str.indexOf(" ");
    int a = Integer.parseInt(str.substring(0,blank));
    int b = Integer.parseInt(str.substring(blank+1));
    long ans = 0;
    if(b==0) {
     bos.write(new Long(a).toString().getBytes());
    }else if (a==0){
     for(int d = 0 ; d <= b ; d++) {
      long temp = dp1[b][d];
      ans += temp; 
     }
     ans %= sp;
     bos.write(new Long(ans).toString().getBytes());
    } else {
     for(int d = 0 ; d < b ; d++) {
      long temp = dp1[b-1][d];
      long mult = 2;
      temp *= mult;
      ans += temp; 
     }
     ans %= sp;
     for(int d = 0 ; d < 32 ; d++) {  
      for(int e = d+1 ; e < 32 ; e++ ) {
       long repValue = list[b][d][e];
       long temp = repValue * Math.min(a+1,pow2[e]-pow2[d]);
       ans += temp;
       ans %= sp;
      }
     }
     ans %= sp;
     long temp = all[b] * (a+1);
     ans += temp;
     ans+=a+2;
     ans %= sp;
     bos.write(new Long(ans).toString().getBytes());
    }
    bos.write(eolb);
   }
   bos.flush();
  } catch(IOException ioe) {
   ioe.printStackTrace();
  }
 }
}









In C :





#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
long long dp[1001][1001][31]={0},dp3[1001];

int main(){
  int T,A,B,l,i,j,k;
  long long t,f;
  for(k=0;k<31;k++)
    for(i=k+1;i<=1000;i++)
      for(j=k+1;j<=1000;j++)
        if(i==k+1){
          if(j==k+1)
            dp[i][j][k]=1;
        }
        else{
          dp[i][j][k]=dp[i-1][j][k];
          if(j>i)
            dp[i][j][k]=(dp[i][j][k]+dp[i-1][j-i][k])%MOD;
          if(i==j)
            dp[i][j][k]=(dp[i][j][k]+1)%MOD;
        }
  scanf("%d",&T);
  while(T--){
    scanf("%d%d",&A,&B);
    i=A;
    for(k=0;i;i>>=1,k++);
    for(i=0,t=1;i<k+1;i++,t<<=1);
    dp3[0]=(A+1)%MOD;
    l=1;
    for(i=f=1,j=2;i<=k && i<=B;i++,j<<=1,l=i)
      if(j+A<t)
        dp3[i]=(j+A+1)%MOD;
      else{
        dp3[i]=t%MOD;
        f=0;
        l=i+1;
        break;
      }
    if(f){
      f=((j>>1)&0x7fffffff);
      for(i=1,j=2;i<=k-1 && i+k<=B;i++,j<<=1,l=i+k)
        if(j+f+A<t)
          dp3[i+k]=(j+f+A+1)%MOD;
        else{
          dp3[i+k]=t%MOD;
          l=i+k+1;
          break;
        }
    }
    for(;l<1001;l++)
      dp3[l]=dp3[l-1];
    for(i=k+1,t=dp3[1000];i<=B;i++)
      t=(t+dp[B][i][k]*dp3[B-i])%MOD;
    printf("%lld\n",(t-1+MOD)%MOD);
  }
  return 0;
}
                        








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