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Commit e562f97

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4 files changed

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4 files changed

+159
-7
lines changed

‎DPandbitmasking/Mehta.cpp‎

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#include <cstdio>
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#include <cmath>
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#include <iostream>
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#include <set>
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#include <algorithm>
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#include <vector>
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#include <map>
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#include <cassert>
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#include <string>
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#include <cstring>
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using namespace std;
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#define rep(i,a,b) for(int i = a; i < b; i++)
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#define S(x) scanf("%d",&x)
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#define P(x) printf("%d\n",x)
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typedef long long int LL;
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typedef pair<int, int > pii;
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const int MAXN = 2001;
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LL P[11] = {1,2,3,5,7,11,13,17,19,23,29};
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LL dp[2][MAXN][MAXN];
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int profit[MAXN];
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int weight[MAXN];
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vector<pii > v;
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int main() {
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int N,W;
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scanf("%d%d",&N,&W);
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assert(N >= 1 && N <= 2000);
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assert(W >= 1 && W <= 2000);
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rep(i,1,N+1) {
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scanf("%d%d",&profit[i], &weight[i]);
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assert(profit[i] >= 1 && profit[i] <= 1000000000);
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assert(weight[i] >= 1 && weight[i] <= 2000);
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v.push_back(make_pair(profit[i], weight[i]));
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}
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sort(v.begin(), v.end());
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rep(i,1,N+1) {
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profit[i] = v[i-1].first;
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weight[i] = v[i-1].second;
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}
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rep(p,0,11) rep(i,1,N+1) rep(j,1,W+1) {
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if(!p) {
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dp[0][i][j] = max(dp[0][i-1][j], dp[0][i][j-1]);
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if(j >= weight[i])
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dp[0][i][j] = max(dp[0][i][j], dp[0][i-1][j-weight[i]]+profit[i]);
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} else {
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int cur = p&1;
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int prev = 1-cur;
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dp[cur][i][j] = dp[prev][i][j];
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dp[cur][i][j] = max(dp[cur][i][j], dp[cur][i-1][j]);
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dp[cur][i][j] = max(dp[cur][i][j], dp[cur][i][j-1]);
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if(j >= weight[i]) {
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dp[cur][i][j] = max(dp[cur][i][j], dp[cur][i-1][j-weight[i]]+profit[i]);
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dp[cur][i][j] = max(dp[cur][i][j], dp[prev][i-1][j-weight[i]]+P[p]*profit[i]);
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}
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}
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}
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cout << dp[0][N][W] << endl;
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return 0;
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}

‎DPandbitmasking/Stringmaker.cpp‎

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/*
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Name: Mehul Chaturvedi
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IIT-Guwahati
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*/
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/*
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PROBLEM STATEMENT
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According to Ancient Ninjas , string making is an art form . There are various methods of string making , one of them uses previously generated strings to make the new one . Suppose you have two strings A and B , to generate a new string C , you can pick a subsequence of characters from A and a subsequence of characters from B and then merge these two subsequences to form the new string.
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Though while merging the two subsequences you can not change the relative order of individual subsequences. What this means is - suppose there two characters Ai and Aj in the subsequence chosen from A , where i < j , then after merging if i acquires position k and j acquires p then k<p should be true and the same apply for subsequence from C.
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Given string A , B , C can you count the number of ways to form string C from the two strings A and B by the method described above. Two ways are different if any of the chosen subsequence is different .
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As the answer could be large so return it after modulo 10^9+7 .
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Input Format :
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Line 1 : String A
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Line 2 : String B
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Line 3 : String C
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Output Format :
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The number of ways to form string C
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Constralls :
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1 <= |A|, |B|, |C| <= 50
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Sample Input :
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abc
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abc
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abc
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Sample Output :
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8
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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#define MOD 1000000007
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ll dp[100][100][100];
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int memsolver(string a,string b,string c){
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ll x = a.size();
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ll y = b.size();
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ll z = c.size();
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ll ans = 0;
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if (z == 0)
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{
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return 1;
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}
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if (x<=0 && y<=0)
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{
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return 0;
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}
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if (dp[x][y][z]!=-1)
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{
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return dp[x][y][z];
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}
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for (ll i = 0; i < a.size(); ++i)
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{
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if (a[i]==c[0])
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{
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ans+=memsolver(a.substr(i+1), b, c.substr(1));
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}
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}
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for (ll i = 0; i < b.size(); ++i)
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{
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if (b[i] == c[0])
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{
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ans+=memsolver(a, b.substr(i+1), c.substr(1));
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}
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}
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dp[x][y][z] = ans%MOD;
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return ans%MOD;
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}
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int solve(string a,string b,string c)
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{
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memset(dp, -1, sizeof(dp));
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return memsolver(a, b, c);
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}
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int main( int argc , char ** argv )
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{
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ios_base::sync_with_stdio(false) ;
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cin.tie(NULL) ;
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string a,b,c;
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cin>>a>>b>>c;
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cout<<solve(a,b,c)<<endl;
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return 0 ;
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}

‎DPandbitmasking/Test.txt‎

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6
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1 1 1 0 1 1
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1 1 1 0 0 1
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0 0 1 0 1 1
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0 1 1 1 0 0
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0 1 0 1 1 0
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0 1 1 0 0 1
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abc
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abc
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abc

‎DPandbitmasking/a.out‎

1.01 KB
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