刁難的李老師(升級版)
Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 67 Solved: 10
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Description
Input
Output
Sample Input
3
2 4 1 5
2 6 2 7
2 8 3 11
Sample Output
1
0
8
HINT
注意!找規律,連續、連續~!定有深層奧義。再次加油!呵呵!~O(∩_∩)O~
思路:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int dp[105];
long long pow_mod(long long a,long long n,long long m)
{
long long ans=1;
while(n)
{
if(n&1) ans=ans*a%m;
a=a*a%m;
n>>=1;
}
return ans;
}
int main()
{
int T;
long long a,b,k,m;
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%lld %lld %lld %lld",&a,&b,&k,&m);
long long ans=0,l=-1;
for(long long i=a;i<=b;i+=2)
{
if(dp[i%m])
{
l=i;
break;
}
dp[i%m]=1;
ans=(ans+pow_mod(i,k,m))%m;
}
if(l==-1) printf("%lld\n",ans);
else
{
long long sum=pow_mod(l,k,m),r=l+2;
for(;;r+=2)
{
if(r%m==l%m) break;
sum=(sum+pow_mod(r,k,m))%m;
}
ans=(ans+(b-l+1)/(r-l)*sum%m)%m;
long long now=(b-l+1)%(r-l);
for(long long i=0;i<=now;i+=2)
ans=(ans+pow_mod(l+i,k,m))%m;
printf("%lld\n",ans);
}
}
return 0;
}