谁是作案嫌疑人？
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
刑侦大队对涉及六个嫌疑人的一桩疑案进行分析：
一、a ，b至少有一人作案；二、a，e,f三人中至少有两人参与作案；三、 a ，d不可能是同案犯；四、b，c或同时作案，或与本案无关；五 c,d中有且只有一人作案；六 如果d没有参与作案则e也不可能参与作案。试编写程序，寻找作案人。Input
多组测试数据，对于每组测试数据，第 1 行输入 6 个空格分隔的整数，代表a、b、c 、d 、e 、f的编号，编号x范围(1 <= x <= 6)，且编号互不相同。
Output
对于每组测试数据，第 1 行至第 6 行分别输出对 a、b、c 、d 、e 、f的判断，详细输出格式请参考样例。
Sample Input
1 2 3 4 5 6
Sample Output
The suspects numbered 1 are criminals.The suspects numbered 2 are criminals.The suspects numbered 3 are criminals.The suspect numbered 4 is not a criminal.The suspect numbered 5 is not a criminal.The suspects numbered 6 are criminals.
#include看着就头疼；#include int main(){ int a,b,c,d,e,f,a1,a2,a3,a4,a5,a6; while(~scanf("%d %d %d %d %d %d",&a1,&a2,&a3,&a4,&a5,&a6)) { int count=0; for(a=0;a<=1;a++) { for(b=0;b<=1;b++) { for(c=0;c<=1;c++) { for(d=0;d<=1;d++) { for(e=0;e<=1;e++) { for(f=0;f<=1;f++) { if(a==1||b==1)count++; if((a==1&&e==1)||(a==1&&f==1)||(e==1&&f==1))count++; if(!(a==1&&d==1))count++; if((b==1&&c==1)||(b==0&&c==0))count++; if((c==0&&d==1)||(c==1&&d==0))count++; if(d==1||(d==0&&e==0))count++; if(count==6) { printf("The suspects numbered %d are criminals.\n",a1); printf("The suspects numbered %d are criminals.\n",a2); printf("The suspects numbered %d are criminals.\n",a3); printf("The suspect numbered %d is not a criminal.\n",a4); printf("The suspect numbered %d is not a criminal.\n",a5); printf("The suspects numbered %d are criminals.\n",a6); } } } } } } } } return 0;}