HDU more than 5,317 schools in the third field

Start test instructions read the wrong shit WA a long time carefully see really water problem one;

It is easy to draw the F () value of each number according to the prime numbers, so you can see that the value of 1 to 1000000 F () is 7 and (2*3*5*7*11*13*17) tells you that the interval only needs to find these 7 numbers in this interval. How many times are there a two-dimensional array of leap "I" "J" (J is less than or equal to 7) indicates the number of J occurrences before I so it's easy to find the value of the interval J. In the record array, you can ask for gcd. Here I am directly listed in all cases!!!

#include <stdio.h> #include <string.h> #include <iostream>using namespaceStd;#define LL (x) (x<<1) #define RR (x) ((x<<1) | |intCont[1000100]={0},Mark[1000010]={0},Leap[1000010][8]={0}; int Max (int A ,int B ) {  return A>B?A:B;} int Main () {  int I,J;Memset(Mark,0 ,sizeof( Mark));Memset(Cont,0 ,sizeof( Cont));  for (I=2;I<=1000000;I + +) {  if( Mark[I ])  continue; Cont[I]=1;  for (J=I+I;J<=1000000;J+=I) {Cont[J]++;Mark[J]=1; }} for ( I=2;I<=1000000;I + +) {  for( J=1;J<=7;J++) {Leap[I][J]=Leap[I-1][J]; }Leap[I][Cont[I]]++; }  int T,L,R;scanf("%d",&T);  while (T--) {scanf("%d%d",&L,&R); int Record[8];Memset(Record,0 ,sizeof( Record));  for (I=1;I<=7;I++) {Record[I]=Leap[R][I]-Leap[L-1][I]; }  int Ans=1;  for (I=1;I<=7;I + +) {  if( Record[I]==0 )  continue; if (Record[I]>1)Ans=Max(Ans,I);  for (J=I+1;J<=7;J + +) {  if( Record[J]==0 )  continue; else if (I==2&&J==4 ) {  if( Ans<2)Ans=2; }  Else if( I==2&&J==6 ) {  if( Ans<2)Ans=2; }  Else if( I==4&&J==6 ) {  if( Ans<2)Ans=2; }  Else if( I==3&&J==6 ) {  if( Ans<3)Ans=3; } } }Printf("%d\n",Ans); }  return 0;} 

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HDU more than 5,317 schools in the third field