[csu1392] YY a bit

Test instructions: Given x, how many 10^8 within the number to satisfy this number multiplied by x, the highest bit to the lowest bit. Set the number of the highest bit and the remaining length, the column equation reasoning can be.

1 #pragmaComment (linker, "/stack:10240000,10240000")2 3#include <iostream>4#include <cstdio>5#include <algorithm>6#include <cstdlib>7#include <cstring>8#include <map>9#include <queue>Ten#include <deque> One#include <cmath> A#include <vector> -#include <ctime> -#include <cctype> the#include <Set> -  - using namespacestd; -  + #defineMem0 (a) memset (a, 0, sizeof (a)) - #defineLson L, M, RT << 1 + #defineRson m + 1, R, RT << 1 | 1 A #definedefine_m int m = (L + r) >> 1 at #defineRep (A, b) for (int a = 0; a < b; a++) - #defineLowbit (x) ((x) & (-(x))) - #defineCONSTRUCTINT4 (name, a, B, C, D) name (int a = 0, int b = 0, int c = 0, int d = 0): A (a), B (b), C (c), D (d) {} - #defineCONSTRUCTINT3 (name, a, B, c) name (int a = 0, int b = 0, int c = 0): A (a), B (b), C (c) {} - #defineConstructInt2 (name, a, b) name (int a = 0, int b = 0): A (a), B (b) {} -  intypedefDoubledb; -typedefLong LongLL; totypedef pair<int,int>PII; +typedef multiset<int>MSI; -typedef multiset<int>:: iterator msii; thetypedefSet<int>si; *typedefSet<int>:: iterator sii; $typedef vector<int>VI;Panax Notoginseng  - Const intdx[8] = {1,0, -1,0,1,1, -1, -1}; the Const intdy[8] = {0, -1,0,1, -1,1,1, -1}; + Const intMAXN = 1e5 +7; A Const intMAXM = 1e5 +7; the Const intMAXV = 1e7 +7; + Const intMD = 1e9 +7; - Const intINF = 1e9 +7; $ Const DoublePI = ACOs (-1.0); $ Const DoubleEPS = 1e-Ten; -  - intDigit (LL x) { the     intCNT =0; -      while(x) {Wuyicnt++; theX/=Ten; -     } Wu     returnCNT; - } About  $ intMain () { -     //freopen ("In.txt", "R", stdin); -     DoubleTX; -      while(Cin >>tx) { ALL x = (int) (TX *10000+0.5),Get=0; +         if(x >=100000) { thePuts"No Solution"); -             Continue; $         } theLL p =1; the          for(inti =0; I <=7; i++) { the              for(intK =1; K <=9; k++) { theLL tmp = k * (x * p-1e4); -                 if(tmp% (LL) (1e5-x))Continue; inTMP/= 1E5-x; the                 if(digit (TMP) = =i) { theprintf"%d", k); About                     if(tmp >0) printf ("%d", TMP); thePuts""); the                     Get=1; the                 } +             } -P *=Ten; the         }Bayi         if(!Get) puts ("No Solution"); the     } the     return 0; -}

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[csu1392] YY a bit