OJ离散数学实验一

1-1 A - 1.1联结词真值运算

#include<stdio.h>
int main()
{
	int p,q;
	while(~scanf("%d %d",&p,&q))
	{
	if(p==1&&q==1) printf("1 ");
	else printf("0 ");
	if(p==0&&q==0) printf("0 ");
	else printf("1 ");
	if(p==1&&q==0) printf("0 ");
	else printf("1 ");
	if(p==q) printf("1 ");
	else printf("0 ");
	if(p==1&&q==1) printf("0 ");
	else printf("1 ");
	if(p==0&&q==0) printf("1\n");
	else printf("0\n");
   }
	return 0; 
}

1-2 B - 1.3按位AND和按位OR

#include<stdio.h>
int main()
{
	int n,a[66],b[66],c[66],d[66],i;
	while(~scanf("%d",&n))
	{
	 for(i=0;i<n;i++)
	 {
		 scanf("%d",&a[i]);
	 }
	 for(i=0;i<n;i++)
	 {
		 scanf("%d",&b[i]);
	 }
	 for(i=0;i<n;i++)
	 {
	 	if(a[i]==1&&b[i]==1) c[i]=1;
	 	else c[i]=0;
	 	if(a[i]==0&&b[i]==0) d[i]=0;
	 	else d[i]=1;
	   }  
	 for(i=0;i<n;i++)
	 {
	 	if(i==n-1)
	 	printf("%d\n",c[i]);
	 	else printf("%d ",c[i]);
	 }
	 for(i=0;i<n;i++)
	 {
	 	if(i==n-1)
	 	printf("%d\n",d[i]);
	 	else printf("%d ",d[i]);
	 }
    }
    return 0;
}

1-3 C - 哪款赛车最佳？

#include<stdio.h>
#include<stdlib.h>
int main()
{
	int a,b,c,d,i,A,B,C,D;
	while(~scanf("%d %d %d",&a,&b,&c))
	{
		for(i=1;i<=4;i++)
		{
			if(a==i)\\如果a为最佳车，下同
			{
				A=1;
				B=0;
				C=1;
				D=1;
			}
			else if(b==i)
			{
				A=0;
				B=1;
				C=1;
				D=0;
			}
			else if(c==i)
			{
				A=0;
				B=0;
				C=0;
				D=1;
			}
			else
			{
				A=0;
				B=0;
				C=1;
				D=1;
			}
			if(A+B+C+D==1)
			{
			if(A==1) printf("%d A\n",i);
			if(B==1) printf("%d B\n",i);
			if(C==1) printf("%d C\n",i);
			if(D==1) printf("%d D\n",i);
			}
		}
    }
    return 0;
}

1-4 D - 谁是作案嫌疑人？

#include<stdio.h>
int main()
{
	int a,b,c,d,e,f,i;
	while(~scanf("%d %d %d %d %d %d",&a,&b,&c,&d,&e,&f))
	{
		printf("The suspects numbered %d are criminals.\n",a);
		printf("The suspects numbered %d are criminals.\n",b);
		printf("The suspects numbered %d are criminals.\n",c);
		printf("The suspect numbered %d is not a criminal.\n",d);
		printf("The suspect numbered %d is not a criminal.\n",e);
		printf("The suspects numbered %d are criminals.\n",f);
	}
	return 0;
}//这个题可以说取巧了，不需要任何计算，因为他们是不是犯罪嫌疑人是已经定了的。

最近在做离散实验，所以浅更新一下，oj程序基础系列还会继续更新的。