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In Topic: Min odd and min even of 6 entry without loop in c/c++
Posted 31 Jan 2012
#include <iostream> #include <cstdio> using namespace std; int main() { int a1,a2,a3,a4,a5,a6; int odd_count=1,even_count=1; int odd1,odd2,odd3,odd4,odd5,odd6; int eve1,eve2,eve3,eve4,eve5,eve6; int min_odd, min_even; printf(" please enter 6 number"); scanf("%d",&a1); scanf("%d",&a2); scanf("%d",&a3); scanf("%d",&a4); scanf("%d",&a5); scanf("%d",&a6); if (a1%2==0) min_even=a1; else if (a1%2!=0) min_odd=a1; if (a2%2==0 && a2<min_even) min_even=a2; else if (a2%2!=0 && a2<min_odd) min_odd=a2; if (a3%2==0 && a3<min_even) min_even=a3; else if (a3%2!=0 && a3<min_odd ) min_odd=a3; if (a4%2==0 && a4<min_even) min_even=a4; else if (a4%2!=0 && a4<min_odd ) min_odd=a4; if (a5%2==0 && a5<min_even) min_even=a5; else if (a5%2!=0 && a5<min_odd ) min_odd=a5; if (a6%2==0 && a6<min_even) min_even=a5; else if (a6%2!=0 && a6<min_odd ) min_odd=a6; printf( "odd %d",min_odd); printf( "odd %d",min_even); system("pause"); }
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