【離散數學一】 誰是作案嫌疑人?
阿新 • • 發佈:2019-02-05
Problem Description
刑偵大隊對涉及六個嫌疑人的一樁疑案進行分析:
一、a ,b至少有一人作案;
二、a,e,f三人中至少有兩人蔘與作案;
三、 a ,d不可能是同案犯;
四、b,c或同時作案,或與本案無關;
五 c,d中有且只有一人作案;
六 如果d沒有參與作案則e也不可能參與作案。
試編寫程式,尋找作案人。
Input
多組測試資料,對於每組測試資料,第 1 行輸入 6 個空格分隔的整數,代表a、b、c 、d 、e 、f的編號,編號x範圍(1 <= x <= 6),且編號互不相同。
Output
對於每組測試資料,第 1 行至第 6 行分別輸出對 a、b、c 、d 、e 、f的判斷,詳細輸出格式請參考樣例。
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.
Hint
gcc解法:#include <stdio.h>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)!=EOF)
{
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;
}
g++解法:
#include <iostream> #include <bits/stdc++.h> using namespace std; int main() { int a,b,c,d,e,f; while(scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f)!=EOF) { 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; }