sicily题目之2001 2005

2001. Scavenger Hunt
	Total: 1099 Accepted: 500 Rating: 1.0/5.0(2 votes)

	Time Limit: 1s Memory Limit: 64MB Description Farmer John has scattered treats for Bessie at special places in the pasture. Since everyone knows that smart cows make tasty milk, FJ has placed the treats at locations that require Bessie to think. He has given her two numbers, P and Q (1 <= P <= 6,000; 1 <= Q <= 6,000), and she has to check every point in the pasture whose x-coordinate is a factor of P and whose y-coordinate is a factor of Q to find her treat. Suppose FJ gives Bessie P = 24 and Q = 2. Here are all of their respective factors: P = 24 => 1, 2, 3, 4, 6, 8, 12, 24 Q = 2 => 1, 2 Bessie would thus check grid locations: (1, 1), (1, 2), (2, 1), (2, 2), (3, 1)... Help Bessie by printing all of the points she ought to check. Input * Line 1: Two space separated integers: P and Q Output * Lines 1..?: A complete list of unique pairs of space-separated integers sorted by the first number and, if tied, the second number: a factor of P followed by a factor of Q Sample Input Copy sample input to clipboard 24 2 Sample Output 1 1 1 2 2 1 2 2 3 1 3 2 4 1 4 2 6 1 6 2 8 1 8 2 12 1 12 2 24 1 24 2

#include<stdio.h>
int main()
{
int P,Q,p[100],q[100],i,j,k,l;
scanf("%d %d",&P,&Q);

for(i=1,j=0;i<=P;i++)
if(P%i==0)
{
p[j]=i;
j++;
}
for(i=1,k=0;i<=Q;i++)
if(Q%i==0)
{
q[k]=i;
k++;
}
for(i=0;i<j;i++)
for(l=0;l<k;l++)
printf("%d %d/n",p[i],q[l]);
}

 

2005. Lovely Number
	Total: 1447 Accepted: 352 Rating: 3.0/5.0(2 votes)

	Time Limit: 1s Memory Limit: 64MB Description   Every time after got a solution from kevin, ivan repeats again and again,"Are there any better ones?". It seems more worse this time. Given a sequence of numbers, which ivan is sure that all of them appear even times except only one "lovely number" appears odd times, kevin must find out the lovely number as soon as possible. So, let's rock! Input   There are multiple test cases. Each test case contains two lines. The first line is an integer N, 1 <= N <= 10001, and N % 2 = 1. The second line contains N integers, separated by spaces. Input is terminated by EOF. Output   For each test case, output a line of an integer, which should be the lovely number. Sample Input Copy sample input to clipboard 3 1 2 1 5 7 7 7 7 3 7 2 2 4 4 4 2 2 Sample Output 2 3 4

#include<stdio.h>
int main()
{
int temp=0,j,k,n,a[10002]={0};
while(1)
{
if(scanf("%d",&n)!=EOF)
{
for(j=0;j<n;j++)
{scanf("%d",&a[j]);}
for(k=0;k<n-1;k++)
for(j=k;j<n-1;j++)
if(a[k]==a[j+1])
{a[k]=0;a[j+1]=0;}
for(k=0;k<n;k++)
temp=temp+a[k];
printf("%d/n",temp);
temp=0;

}

else
break;
}


return 0;
}