POJ 3982
发布日期:2021-06-30 15:30:56
浏览次数:2
分类:技术文章
本文共 5694 字,大约阅读时间需要 18 分钟。
序列
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8557 | Accepted: 3883 |
Description
数列A满足An = An-1 + An-2 + An-3, n >= 3
编写程序,给定A0, A1 和 A2, 计算A99Input
输入包含多行数据
每行数据包含3个整数A0, A1, A2 (0 <= A0, A1, A2 <= 32767) 数据以EOF结束Output
对于输入的每一行输出A99的值
Sample Input
1 1 1
Sample Output
69087442470169316923566147
代码:
JAVA
import java.util.Scanner; import java.math.*; public class Main{ public static void main(String[] args){ Scanner cin = new Scanner(System.in); BigInteger a, b, c, ans, temp1, temp2; while(cin.hasNextBigInteger()){ a = cin.nextBigInteger(); b = cin.nextBigInteger(); c = cin.nextBigInteger(); ans = c; int i = 3; while(i <= 99){ temp1 = a.add(b); ans = temp1.add(c); a = b; b = c; c = ans; /*System.out.println(a); System.out.println(b); System.out.println(c); System.out.println(temp1); System.out.println(temp2);*/ i++; } System.out.println(ans); } } }
C++
#include#include #include #include #include using namespace std; const int MAXN =510; struct bign{ int len, s[MAXN]; bign () { memset(s, 0, sizeof(s)); len = 1; } bign (int num) { *this = num; } bign (const char *num) { *this = num; } bign operator = (const int num) { char s[MAXN]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { for(int i = 0; num[i] == '0'; num++) ; //去前导0 len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } bign operator + (const bign &b) const //+ { bign c; c.len = 0; for(int i = 0, g = 0; g || i < max(len, b.len); i++) { int x = g; if(i < len) x += s[i]; if(i < b.len) x += b.s[i]; c.s[c.len++] = x % 10; g = x / 10; } return c; } bign operator += (const bign &b) { *this = *this + b; return *this; } void clean() { while(len > 1 && !s[len-1]) len--; } bign operator * (const bign &b) //* { bign c; c.len = len + b.len; for(int i = 0; i < len; i++) { for(int j = 0; j < b.len; j++) { c.s[i+j] += s[i] * b.s[j]; } } for(int i = 0; i < c.len; i++) { c.s[i+1] += c.s[i]/10; c.s[i] %= 10; } c.clean(); return c; } bign operator *= (const bign &b) { *this = *this * b; return *this; } bign operator - (const bign &b) { bign c; c.len = 0; for(int i = 0, g = 0; i < len; i++) { int x = s[i] - g; if(i < b.len) x -= b.s[i]; if(x >= 0) g = 0; else { g = 1; x += 10; } c.s[c.len++] = x; } c.clean(); return c; } bign operator -= (const bign &b) { *this = *this - b; return *this; } bign operator / (const bign &b) { bign c, f = 0; for(int i = len-1; i >= 0; i--) { f = f*10; f.s[0] = s[i]; while(f >= b) { f -= b; c.s[i]++; } } c.len = len; c.clean(); return c; } bign operator /= (const bign &b) { *this = *this / b; return *this; } bign operator % (const bign &b) { bign r = *this / b; r = *this - r*b; return r; } bign operator %= (const bign &b) { *this = *this % b; return *this; } bool operator < (const bign &b) { if(len != b.len) return len < b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] < b.s[i]; } return false; } bool operator > (const bign &b) { if(len != b.len) return len > b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] > b.s[i]; } return false; } bool operator == (const bign &b) { return !(*this > b) && !(*this < b); } bool operator != (const bign &b) { return !(*this == b); } bool operator <= (const bign &b) { return *this < b || *this == b; } bool operator >= (const bign &b) { return *this > b || *this == b; } string str() const { string res = ""; for(int i = 0; i < len; i++) res = char(s[i]+'0') + res; return res; }}; istream& operator >> (istream &in, bign &x){ string s; in >> s; x = s.c_str(); return in;} ostream& operator << (ostream &out, const bign &x){ out << x.str(); return out;}//大数模板 struct matrix//构建矩阵{ bign f[4][4]; matrix operator *( matrix &a) { matrix res; for (int i=1;i<=3;i++) { for (int j=1;j<=3;j++) { res.f[i][j]=0; for (int k=1;k<=3;k++) res.f[i][j]+=f[i][k]*a.f[k][j]; } } return res; }}a,b;int a0,a1,a2;void init(){ b.f[1][1]=b.f[1][2]=b.f[1][3]=1; b.f[2][1]=1; b.f[3][2]=1; a.f[1][1]=a2,a.f[2][1]=a1,a.f[3][1]=a0;} matrix fast_pow(matrix base,int k)//快速幂{ matrix ans; for (int i=1;i<=3;i++) ans.f[i][i]=1; while (k) { if (k&1) ans=ans*base; base=base*base; k>>=1; } return ans;}int main(){ while (scanf("%d%d%d",&a0,&a1,&a2)!=EOF) { init(); matrix cur,ans; cur=b; cur=fast_pow(cur,97); ans=cur*a; cout< <
转载地址:https://joycez.blog.csdn.net/article/details/82824644 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!
发表评论
最新留言
感谢大佬
[***.8.128.20]2024年04月10日 01时55分51秒
关于作者
喝酒易醉,品茶养心,人生如梦,品茶悟道,何以解忧?唯有杜康!
-- 愿君每日到此一游!
推荐文章
Request_继承体系
2019-05-01
Request_获取请求行数据_方法介绍
2019-05-01
spring的优势
2019-05-01
编写jdbc的工程代码用于分析程序的耦合
2019-05-01
工厂模式解耦的升级版
2019-05-01
hrm中的jwt认证:获取用户数据
2019-05-01
前端权限控制:获取用户信息接口构造数据
2019-05-01
前端权限控制:前端菜单控制
2019-05-01
前端权限控制:实现思路分析-待修改
2019-05-01
有状态服务和无状态服务
2019-05-01
基于jwt的用户鉴权:配置拦截器并测试
2019-05-01
POI的概述
2019-05-01
POI文件导入:跨服务器调用查询部门信息
2019-05-01
DataURL:概述
2019-05-01
DataURL:实现原理及优缺点分析
2019-05-01
DataURL:实现员工头像保存
2019-05-01
DataURL:员工头像回显
2019-05-01
七牛云存储:通过SDK上传图片
2019-05-01
七牛云存储:断点续传
2019-05-01
七牛云存储:实现员工头像保存
2019-05-01