implementation of sieve's algorithm

C-C++/implementation of sieve_algorithm.cpp

@@ -0,0 +1,85 @@
+/*
+All submissions for this problem are available.Prime numbers are arranged in a ordered list U, in increasing order. Let S be a sublist of U with a unique property that for every element A belonging to list S, if i denotes the index of A in list U, than i also belongs to list U.
+
+Given N, find sum of first N elements of list S, assuming 1-based indexing.
+
+As the sum can be very large, print the sum modulo 109+7.
+
+Input:
+-The first line of the input contains a single integer T denoting the number of test cases.
+
+-Only line of each test case has an integer N .
+
+Output:
+For each test case, print a single integer denoting the sum of first N elements of set S modulo 109+7.
+
+Constraints
+1=T=10000
+1=N=1000*/
+#include<iostream>
+#include<vector>
+#include<cmath>
+using namespace std;
+int v[100000]={0};
+int su[100000]={0};
+int prime[100000];
+void calculate()
+{
+    int i,j;
+       int sum;
+    v[0]=1;
+    	v[1]=1;
+    	su[0]=0;
+		for(i=2;i<sqrt(100000);i++)
+   if(v[i]==false)
+     for (j = pow(i,2);j<=100000;j+=i)
+       v[j] = true;
+       sum=0;
+       int k=0;
+       prime[0] = 0;
+       for(i=1;i<=100000;i++)
+       {
+       	//king=0;
+           if(v[i]==0)
+           {
+		   k++;
+           	prime[k] = i;
+           	//cout<<i;
+		   }
+           //v[]
+       }
+       int lo;
+       su[0] = 0;
+       lo=1;
+       for(i=1;i<=k;i++)
+       {
+       	if(v[i]==0)
+       	{
+       		sum = sum+prime[i];
+       		su[lo] = sum;
+       		lo++;
+		   }
+	   }
+}
+int main()
+{
+std::ios::sync_with_stdio(false);
+//cin.tie(0);
+
+	int t;
+	long int n;
+	calculate();
+    cin>>t;
+
+   // vector<int> v;
+
+    while(t--)
+    {
+    	cin>>n;
+    //	cout<<prime[n];
+
+       cout<<su[n]<<endl;
+       //v.clear();
+	}
+	return 0;
+}