# SJTU OJ 1201. Superxor

Source: Internet
Author: User

Description

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

``  1001+ 1101-------  0100``

Everything is a addition, except that there is no carries.

After realizing this, Pangzi invented Super XOR. It is just a 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 <=), indicating the number of test cases.

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

Output Format

Output T lines, each of which is a integer:the answer for the corresponding test case.

Sample Input
``51234120001``
Sample Output
``1360240001``
Case Limits

Time limit:500 msec

Memory limit:64 MB

I read the wrong question at first. There's no one who can read it wrong.

is to ask 1~n these numbers and not carry them. Not very difficult. But attention to detail, the idea must be clear. Otherwise infinite tle infinite re

Idea: Hit the table found at the end is 99 of the result is 0 then find than n small end is 99 of the ***99~n to ask for it and on the line.

`/************************************************author:guanjuncreated TIME:2016/3/12 9:39:27File name:s jtu1201.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>#definell Long Long#defineMoD 90001#defineINF 0x3f3f3f3f#defineMAXN 10010#defineCLE (a) memset (A,0,sizeof (a))Const Doubleeps=1e-5;using namespacestd;intar[ -],num;voidAdd (ll x) {num=0;  while(x>0) {Ar[num]+ = (xTen); Ar[num]%=Ten; X/=Ten; Num++; }}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin); #endif    //freopen ("OUT.txt", "w", stdout);    intT;    ll N,k,m; CIN>>T;  while(t--) {cin>>N; M=N;        CLE (AR); if(n> About){             while(1){                if(n% -== About) {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(intI=0; i<num;i++) {ans+=ar[i]*tmp; TMP*=Ten; } cout<<ans<<Endl; }    return 0;}`

SJTU OJ 1201. Superxor

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