zoukankan      html  css  js  c++  java
  • POJ

    http://poj.org/problem?id=1503
    直接抄模板就可以了。

    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<map>
    #include<set>
    #include<stack>
    #include<string>
    #include<queue>
    #include<vector>
    using namespace std;
    typedef long long ll;
    
    struct BigInt {
        const static int BASE = 1000000000;
        const static int DLEN = 9;
        vector<int> a;
        int len;
    
        BigInt() {
            a.resize(4);
            len = 1;
        }
    
        BigInt(int v) {
            a.resize(4);
            len = 0;
            do {
                a[len++] = v % BASE;
                v /= BASE;
            } while(v);
        }
    
        BigInt(const char *s) {
            int L = strlen(s);
            len = L / DLEN;
            if(L % DLEN)
                len++;
            a.resize(len + 1);
            int id = 0;
            for(int i = L - 1; i >= 0; i -= DLEN) {
                int t = 0;
                int k = i - DLEN + 1;
                if(k < 0)
                    k = 0;
                for(int j = k; j <= i; j++)
                    t = t * 10 + s[j] - '0';
                a[id++] = t;
            }
        }
    
        BigInt operator +(const BigInt &b)const {
            BigInt res;
            res.len = max(len, b.len);
            res.a.resize(res.len + 1);
            for(int i = 0; i < res.len; i++) {
                res.a[i] += ((i < len) ? a[i] : 0) + ((i < b.len) ? b.a[i] : 0);
                res.a[i + 1] += res.a[i] / BASE;
                res.a[i] %= BASE;
            }
            if(res.a[res.len] > 0)
                res.len++;
            return res;
        }
    
        BigInt operator *(const BigInt &b)const {
            //高精乘高精,考虑用FFT优化
            BigInt res;
            res.a.resize(len + b.len);
            for(int i = 0; i < len; i++) {
                int up = 0;
                for(int j = 0; j < b.len; j++) {
                    ll temp = 1ll * a[i] * b.a[j] + res.a[i + j] + up;
                    res.a[i + j] = temp % BASE;
                    up = temp / BASE;
                }
                if(up != 0)
                    res.a[i + b.len] = up;
            }
            res.len = len + b.len;
            while(res.a[res.len - 1] == 0 && res.len > 1)
                res.len--;
            return res;
        }
    
        BigInt operator *(const int &b)const {
            //高精乘低精
            BigInt res;
            res.a.resize(len + 1);
            for(int i = 0; i < len; i++) {
                int up = 0;
                ll temp = 1ll * a[i] * b + res.a[i] + up;
                res.a[i] = temp % BASE;
                up = temp / BASE;
                if(up != 0)
                    res.a[i + 1] = up;
            }
            res.len = len + 1;
            while(res.a[res.len - 1] == 0 && res.len > 1)
                res.len--;
            return res;
        }
    
        bool operator <(const BigInt &b)const {
            if(len != b.len)
                return len < b.len;
            else {
                for(int ln = len - 1; ln >= 0; ln--) {
                    if(a[ln] != b.a[ln])
                        return a[ln] < b.a[ln];
                }
                return false;
            }
        }
    
        void output() {
            printf("%d", a[len - 1]);
            for(int i = len - 2; i >= 0 ; i--)
                printf("%09d", a[i]);
            printf("
    ");
        }
    };
    
    char s[205];
    int main() {
    #ifdef Yinku
        freopen("Yinku.in", "r", stdin);
    #endif // Yinku
        BigInt ans;
        while(~scanf("%s", s)) {
            int n = strlen(s);
            if(n == 1 && s[0] == '0')
                break;
            ans = ans + BigInt(s);
        }
        ans.output();
        return 0;
    }
    
  • 相关阅读:
    第二节,神经网络中反向传播四个基本公式证明——BackPropagation
    第一节,windows和ubuntu下深度学习theano环境搭建
    oracle和SQLserver数据库中select into 的区别
    Mysql与Oracle区别
    SQLserver 设置自增为显式插入
    SQL 存储过程入门(事务)(四)
    SQLSqlserver中如何将一列数据,不重复的拼接成一个字符串
    SQL命令优化(积累)
    手机游戏运营主要的指标是什么? 7天活跃, 14天活跃 ARPU ?如何提升游戏 app 的虚拟道具的收入?
    从用户心理看游戏运营和推广
  • 原文地址:https://www.cnblogs.com/Inko/p/11715112.html
Copyright © 2011-2022 走看看