UVa 10375 - Choose and divide

简单的分解质因数约分

  1 #include <cstdio>

  2 #include <cstdlib>

  3 #include <cstring>

  4 

  5 const int MAXN = 10000 + 10;

  6 

  7 bool Prime[MAXN];

  8 

  9 int aaa[5];

 10 int bbb[5];

 11 int num[2][MAXN];

 12 

 13 void GetPrime()

 14 {

 15     memset( Prime, true, sizeof(Prime) );

 16     Prime[0] = false ;

 17     Prime[1] = false ;

 18     for ( int i = 2; i < MAXN; i++ )

 19         if ( Prime[i] )

 20         {

 21             for ( int j = i * i; j < MAXN; j += i )

 22                 Prime[j] = false;

 23         }

 24     return;

 25 }

 26 

 27 int cmp( const void *a, const void *b )

 28 {

 29     return *(int *)a - *(int *)b;

 30 }

 31 

 32 int max( int a, int b )

 33 {

 34     return a > b ? a : b;

 35 }

 36 

 37 int min( int a, int b )

 38 {

 39     return a < b ? a : b;

 40 }

 41 

 42 void show()

 43 {

 44     for ( int i = 0; i < 20; i++ )

 45        printf("%d ", num[0][i]);

 46     puts("\n---------------------");

 47     for ( int i = 0; i < 20; i++ )

 48        printf("%d ", num[1][i]);

 49     puts("\n----------------");

 50     return;

 51 }

 52 

 53 int main()

 54 {

 55     GetPrime();

 56     int p, q, r, s;

 57     while ( scanf( "%d%d%d%d", &p, &q, &r, &s ) != EOF )

 58     {

 59         memset( num, 0, sizeof(num) );

 60 

 61         aaa[0] = p;

 62         aaa[1] = s;

 63         aaa[2] = r - s;

 64         bbb[0] = q;

 65         bbb[1] = r;

 66         bbb[2] = p - q;

 67         qsort( aaa, 3, sizeof(int), cmp );

 68         qsort( bbb, 3, sizeof(int), cmp );

 69 

 70 //        for ( int i = 0; i < 3; i++ ) printf("%d %d\n", aaa[i], bbb[i]);

 71 

 72         int down, up;

 73         for ( int i = 0; i < 3; i++ )

 74         {

 75             int xxx;

 76             if ( aaa[i] == bbb[i] ) continue;

 77             else if ( aaa[i] > bbb[i] )

 78             {

 79                 down = bbb[i] + 1;

 80                 up = aaa[i];

 81                 xxx = 0;

 82             }

 83             else

 84             {

 85                 down = aaa[i] + 1;

 86                 up = bbb[i];

 87                 xxx = 1;

 88             }

 89             for ( int i = down; i <= up; i++ )

 90             {

 91                 int tp = i;

 92     //            printf("tp = %d\n", tp);

 93                 int yyy = 2;

 94                 while ( tp != 1 )

 95                 {

 96                     if ( Prime[yyy] && tp % yyy == 0 )

 97                     {

 98       //                  printf("yyy = %d\n", yyy);

 99                         tp /= yyy;

100                         ++num[xxx][yyy];

101                     }

102                     else ++yyy;

103                 }

104             }

105         }

106 

107         for ( int i = 2; i < MAXN; i++ )

108         if ( Prime[i] )

109         {

110             if ( num[0][i] > num[1][i] )

111             {

112                 num[0][i] -= num[1][i];

113                 num[1][i] = 0;

114             }

115             else

116             {

117                 num[1][i] -= num[0][i];

118                 num[0][i] = 0;

119             }

120         }

121 

122   //      show();

123 

124         double ans = 1.0;

125         for ( int i = 2; i < MAXN; i++ )

126             if ( Prime[i] )

127             {

128                 int fenzi = 1, fenmu = 1;

129                 for ( int j = 0; j < num[0][i]; j++ ) fenzi *= i;

130                 for ( int j = 0; j < num[1][i]; j++ ) fenmu *= i;

131                 ans *= (double)fenzi/fenmu;

132             }

133 

134         printf( "%.5lf\n", ans );

135     }

136     return 0;

137 }