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#include<bits/stdc++.h>usingnamespacestd;stringltrim(conststring&);stringrtrim(conststring&);/* * Complete the 'summingSeries' function below. * * The function is expected to return an INTEGER. * The function accepts LONG_INTEGER n as parameter. *//*T_n=n^2-(n-1)^2a^2+b^2=(a+b)(a-b)T_n = (n+n-1)(n-(n-1))=(2n-1)(1)=2n-1S_n = T_1+T_2+...+T_nS_n = sum_{k=1}^{n} (2k-1)S_n = 2sum_{k=1}^{n} k - sum_{k=1}^{n} 1S_n = 2* (n*n-1)/2 -n = n(n+1)-n = n^2==> S_n = n^2S_n mod (10^9+7) = n^2 mod 1000000007*/intsummingSeries(longn){// return (1LL*n*n)%1000000007; // Fail 2 testcasesconstintMOD=1000000007;n%=MOD;return(n*n)%MOD;}intmain(){ofstreamfout(getenv("OUTPUT_PATH"));stringt_temp;getline(cin,t_temp);intt=stoi(ltrim(rtrim(t_temp)));for(intt_itr=0;t_itr<t;t_itr++){stringn_temp;getline(cin,n_temp);longn=stol(ltrim(rtrim(n_temp)));intresult=summingSeries(n);fout<<result<<"\n";}fout.close();return0;}stringltrim(conststring&str){strings(str);s.erase(s.begin(),find_if(s.begin(),s.end(),not1(ptr_fun<int,int>(isspace))));returns;}stringrtrim(conststring&str){strings(str);s.erase(find_if(s.rbegin(),s.rend(),not1(ptr_fun<int,int>(isspace))).base(),s.end());returns;}
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Summing the N series
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Here's my solution: