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  • sjtu oj 1201. SuperXOR

    Description

    Pangzi recently realized that bitwise XOR operation is just an addition without carries. For example, when computing (1001)_2 XOR (1101)_2, you write down:

      1001
    + 1101
    -------
      0100
    

    You see, everything is like an addition, except that there are no carries.

    After realizing this, Pangzi invented Super XOR. It is just an addition on decimal numbers without carries. For example, 123 SUPERXOR 789 = 802, since 1+7=8, 2+8=0, 3+9=2.

    Now comes a question. Given N, find 1 SUPERXOR 2 SUPERXOR 3 SUPERXOR 4 SUPERXOR ... SUPERXOR N

    Input Format

    The first line contains an integer T (1 <= T <= 1000), indicating the number of test cases.

    T lines follow each containing 1 integers N (1 <= N <= 10^12).

    Output Format

    Output T lines, each line is an integer: the answer for the corresponding test case.

    Sample Input

    5
    1
    2
    3
    4
    120001
    

    Sample Output

    1
    3
    6
    0
    240001
    

    Case Limits

    Time limit: 500 msec

    Memory limit: 64 MB

    一开始读错题了..这都能读错 也是没谁了。

    就是求1~n这些数的和 不进位。难度不大。但注意细节,思路一定要清晰。否则无限tle  无限re

    思路:打表发现末尾是99的 结果是0 那么找到比n小的末尾是99的 对***99~n 求其和就行了。

    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/3/12 9:39:27
    File Name     :sjtu1201.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ll long long
    
    #define mod 90001
    #define INF 0x3f3f3f3f
    #define maxn 10010
    #define cle(a) memset(a,0,sizeof(a))
    
    const double eps=1e-5;
    using namespace std;
    
    int ar[100],num;
    void add(ll x){
        num=0;
        while(x>0){
            ar[num]+=(x%10);
            ar[num]%=10;
            x/=10;
            num++;
        }
    }
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int t;
        ll n,k,m;
        cin>>t;
        while(t--){
            cin>>n;
            m=n;
            cle(ar);
            if(n>99){
                while(1){
                    if(n%100==99){
                        k=n+1;
                        break;
                    }
                    n--;
                }
                for(ll i=k;i<=m;i++)add(i);
            }
            else for(ll i=1;i<=n;i++)add(i);
            ll ans=0;
            ll tmp=1;
            for(int i=0;i<num;i++){
                ans+=ar[i]*tmp;
                tmp*=10;
            }
            cout<<ans<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/5268540.html
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