BBP算法板子

来源：互联网发布：java装饰模式和注解编辑：程序博客网时间：2024/06/05 05:32

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6217
百度BBP解释：https://wenku.baidu.com/view/716dbac708a1284ac85043bf.html
板子，不会，别问我

#include <cstdio>#include<iostream>using namespace std;typedef long long ll;ll remain( ll m, ll n, ll k, ll extra){    ll temp1=1,temp2=1;    if(n==0)return extra%k;    if(n==1)return (m*extra)%k;    while(n>1)    {        temp1=m;        temp1*=temp1;        if(temp1>=k)temp1%=k;        if(n%2==1)temp2=m*temp2;        temp2%=k;        m=temp1;        n=n/2;    }    temp1=(temp1*temp2)%k;    return (temp1*extra)%k;} ll remain_nex( ll m, ll n, ll k){     ll temp1=1,temp2=1;    if(n==0)return 1;    if(n==1)return m%k;    while(n>1)    {        temp1=m;        temp1*=temp1;        if(temp1>=k)temp1%=k;        if(n%2==1)temp2=m*temp2;        temp2%=k;        m=temp1;        n=n/2;    }    return (temp1*temp2)%k;}char compute_n(int j){     ll m;    long double sum=0,temp=1.0,temp1;    int i;    j--;    temp1=1.0;    for(i=0;i<=j;i++)sum=sum+remain(16,j-i,8*i+1,4)/(long double)(8.0*i+1);    for(i=0;i<=j;i++)sum=sum-remain(16,j-i,8*i+4,2)/(long double)(8.0*i+4);    for(i=0;i<=j;i++)sum=sum-remain_nex(16,j-i,8*i+5)/(long double)(8.0*i+5);    for(i=0;i<=j;i++)sum=sum-remain_nex(16,j-i,8*i+6)/(long double)(8.0*i+6);    temp=1.0;    for(;temp>0.000001;i++)    {        temp=temp/16.0;sum=sum+(4.0/(8*i+1)-2.0/(8*i+4)-1.0/(8*i+5)-1.0/(8*i+6))*temp;    }    for(;sum<0;)sum=sum+16;    m=sum;    sum=sum-m;    sum=sum*16;    m=sum;    return (char)(m<10 ? m+48: m+55);}int main(){    int T, ca, i, n;    scanf("%d", &T);    for (ca = 1;ca <= T;ca++) {        scanf("%d", &n);        printf("Case #%d: %d %c\n", ca, n, compute_n(n));    }}

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