zoukankan      html  css  js  c++  java
  • csuoj-1720-How to Get 2^n

    题目:

    Description

    Johnson is very interested in the k-th power of 2 number, such as 2,4,8,16 ....

    So he always boasts that he is good at solving problems with the power of two.

    One day, Alice could not stand Johnson’s bragging, and she come up a question to heckle him,

    Alice gave Johnson n numbers, and let Johnson takes two numbers inside guaranteeing sum of them is power of 2. Finally Alice let Johnson tell her how many methods can he get two valid numbers.

    Because Alice is goddess in Johnson’s heart , so he do not want to lose face in front of her . But he is very tight facing to his goddess , he hope you can help him solve this problem

    Input

    The first line contains an integer T (T<=2), means there are T test cases.

    For each case

    The first line contains an integer n, means there are n numbers.

    (1<=n<=100000)

    The next line contains n integers(v0,v1.....vn-1) (1<=vi<=10^30 , 2^100 approximately equal 10^30)

    Output

    Each case output the number of method.

    Sample Input

    1
    5
    3 2 5 4 1

    Sample Output

    2
    分析:
    1,打表求出2的K次方(k从1到100);
    2,输入数据,将大数从小到大排序,合并相同的数,并记录相同的个数;
    3,对于每个大数,分别对2的k次方用二分法求出对应的另一个大数,如果相同,则加上该数数目 * (该数数目 - 1) / 2,否则加上两数数目乘积。

    代码:
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int goal[100][31];
    struct BigNum{
        long long cnt;
        int num[31];
    }bigNum[100000];
    
    int cmp(const struct BigNum &a,const struct BigNum &b){
        for(int i = 0;i < 31;i++){
            if(a.num[i] > b.num[i]) return 0;
            else if(a.num[i] < b.num[i]) return 1;
        }
        return 1;
    }
    
    void init(){
        goal[0][30] = 2;
        for(int i = 1;i < 100;i++){
            for(int j = 30;j >= 0;j--){
                int x = goal[i - 1][j] * 2;
                goal[i][j] += x % 10;
                if(j > 0) goal[i][j - 1] = x / 10;
            }
        }
    }
    
    int same(int *a,int *b){
        for(int i = 0;i < 31;i++){
            if(a[i] > b[i]) return 1;
            else if(a[i] < b[i]) return -1;
        }
        return 0;
    }
    
    int find(int *ar,struct BigNum *temp,int b,int cnt){
        int begin = b;
        int end = cnt - 1;
        int middle = (begin + end) / 2;
        while(begin <= end){
            int result = same(ar,temp[middle].num);
            if(result == 0) return middle;
            else if(result == 1){
                begin = middle + 1;
                middle = (begin + end) / 2;
            }
            else{
                end = middle - 1;
                middle = (begin + end) / 2;
            }
        }
        return -1;
    }
    
    int main(){
        int t;
        cin >> t;
        init();
        while(t--){
            int n;
            cin >> n;
            for(int i = 0;i < n;i++){
                char ch[32];
                for(int j = 0;j <= 31;j++) ch[j] = '';
                cin >> ch;
                int cnt = 0;
                for(int j = 0;j <= 31;j++){
                    if(ch[j] != '') cnt++;
                    else break;
                }
                int index = 0;
                for(int j = 31 - cnt;j <= 30;j++) bigNum[i].num[j] = ch[index++] - '0';
                for(int j = 0;j < 31 - cnt;j++) bigNum[i].num[j] = 0;
            }
            sort(bigNum,bigNum + n,cmp);
            for(int i = 0;i < n;i++) bigNum[i].cnt = 1;
            struct BigNum *temp = new struct BigNum[n];
            int cnt = 0;
            int last[31];
            for(int i = 0;i < 31;i++) temp[cnt].num[i] = last[i] = bigNum[0].num[i];
            temp[cnt].cnt = 1;
            cnt++;
            for(int i = 1;i < n;i++){
                int result = same(last,bigNum[i].num);
                if(result == 0) {
                    temp[cnt - 1].cnt++;
                }
                else{
                    for(int j = 0;j < 31;j++) last[j] = temp[cnt].num[j] = bigNum[i].num[j];
                    temp[cnt].cnt = 1;
                    cnt++;
                }
            }
            long long res = 0;
            for(int i = 0;i < cnt;i++){
                for(int j = 0;j < 100;j++){
                    int flag = same(temp[i].num,goal[j]);
                    if(flag == -1){
                        int big[31];
                        int chu[31];
                        for(int k = 0;k < 31;k++) chu[k] = 0;
                        for(int k = 0;k < 31;k++) big[k] = goal[j][k];
                        for(int k = 30;k >= 0;k--){
                            if(big[k] >= temp[i].num[k])
                                chu[k] = big[k] - temp[i].num[k];
                            else{
                                chu[k] = big[k] + 10 - temp[i].num[k];
                                big[k - 1]--;
                            }
                        }
                        int index = find(chu,temp,i,cnt);
                        if(index != -1){
                            if(index == i) res += temp[index].cnt * (temp[index].cnt - 1) / 2;
                            else res += temp[index].cnt * temp[i].cnt;
                        }
                    }
                }
            }
            cout << res << endl;
        }
        return 0;
    }
     
  • 相关阅读:
    go爬虫
    node简单爬虫request简单运用
    Centos7.4安装Mysql5.6
    HTML本地资源读取!
    node-request模块
    react中使用AntDesign库 --- babel-plugin-import 配置
    Django 模型(数据库)
    TypeScript--安装依赖,vscode配置ts自动转换成js文件
    python爬虫
    nodejs爬虫简单实现
  • 原文地址:https://www.cnblogs.com/tracy520/p/6706978.html
Copyright © 2011-2022 走看看