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### icode_8990's blog

By icode_8990, history, 4 weeks ago, ,

Hello, Newbie here. I am trying to solve this problem and this is the solution i am referring to:

#include<bits/stdc++.h>
using namespace std;
#define mod 1000000007
long long num[100010],pw[100010];
int main()
{
long long n,q,i,res,l,r;
string s;
cin>>n>>q;
cin>>s;
for(i=0;i<n;i++)
{
num[i+1] =num[i]+(s[i]=='0');
}
pw[0]=1;
for(i=1;i<n+5;i++)
{
pw[i] = pw[i-1]*2;
if(pw[i]>=mod) pw[i]-=mod;
}
while(q--)
{
scanf("%lld%lld",&l,&r);
res=(mod+pw[r-l+1]-pw[num[r]-num[l-1]])%mod;
printf("%lld\n",res);
}

return 0;
}


I am trying to understand the modulo used here. why does he add another mod in res. How does it help?

Thanks!

•
• +3
•

 » 4 weeks ago, # |   0 It's so that if it goes negative, it will be rectified.
 » 4 weeks ago, # |   0 C++ % operator is undefined behavior with negative number (depends on compiler). Adding mod keeps it positive while maintaining the answer.
•  » » 4 weeks ago, # ^ |   0 no, it's defined behavior
•  » » » 4 weeks ago, # ^ |   0 I mean to say sign is implementation defined.
•  » » » » 4 weeks ago, # ^ | ← Rev. 2 →   0 Since C++11 the sign of the result is well defined, not depending on implementation, see https://stackoverflow.com/questions/13100711/operator-modulo-change-in-c-11