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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1286 Accepted Submission(s): 515
Total Submission(s): 1286 Accepted Submission(s): 515
Problem Description
Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
const int MAX=100000*2;
const int INF=1e9;
int main()
{
int n,m,ans,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(i=1;i<=n;i++)
{
if(i&1)ans=(ans*2+1)%m;
else ans=ans*2%m;
}
printf("%d\n",ans);
}
return 0;
}
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
const int MAX=100000*2;
const int INF=1e9;
int main()
{
int n,m,ans,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(i=1;i<=n;i++)
{
if(i&1)ans=(ans*2+1)%m;
else ans=ans*2%m;
}
printf("%d\n",ans);
}
return 0;
}
Input
Multi test cases,each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000
[Technical Specification]
1<=n, m <= 1000000000
Output
For each case,output an integer,represents the output of above program.
Sample Input
1 10 3 100
AC C++(网上竟然暂时没见到用快速幂矩阵做这道题的程序,然后我自己写了一个)
#include<iostream>
#include<cstring>
#include<cstdio>
//#define DEBUG
struct Matrix
{
long long m[2][2];
Matrix()
{
memset(m,0,sizeof(m));
}
};
long long n,m;
Matrix mul(Matrix a,Matrix b,int m)
{
Matrix res;
for(long long i=0;i<2;++i)
for(long long j=0;j<2;j++)
{
for(long long k=0;k<2;k++)
{
res.m[i][j]+=a.m[i][k]*b.m[k][j];
}
res.m[i][j]%=m;
}
return res;
}
int main()
{
Matrix x,a,res;
a.m[0][0]=1;
a.m[0][1]=0;
a.m[1][0]=1;
a.m[1][1]=2;
while(scanf("%lld%lld",&n,&m)!=EOF)
{
x.m[0][0]=1;
x.m[0][1]=0;
x.m[1][0]=2;
x.m[1][1]=4;
res.m[0][0]=1;
res.m[0][1]=0;
res.m[1][0]=0;
res.m[1][1]=1;
if(n&1)
{
n=(n-1)/2;
while(n)
{
if(n&1) res=mul(x,res,m);
x=mul(x,x,m);
n>>=1;
}
res=mul(a,res,m);
#ifdef DEBUG
for(long long i=0;i<2;++i)
{
for(long long j=0;j<2;++j)
std::cout<<res.m[i][j]<<' ';
std::cout<<std::endl;
}
#endif
}else
{
n=n/2;
while(n)
{
if(n&1) res=mul(x,res,m);
x=mul(x,x,m);
n>>=1;
}
#ifdef DEBUG
for(long long i=0;i<2;++i)
{
for(long long j=0;j<2;++j)
std::cout<<res.m[i][j]<<' ';
std::cout<<std::endl;
}
#endif
}
std ::cout<<res.m[1][0]<<std::endl;
}
return 0;
}