Light OJ 1116 - Ekka Dokka (简单数学)
2016-08-13 11:52
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1116 - Ekka Dokka
Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the
cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive
integers.
They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.
Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.
print the result where M is as small as possible.
思路:奇数乘偶数必定为偶数,所以w为奇数时 直接不符合题意;
w为偶数时 1*w==w所以必有解,遍历一下
代码:
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive
integers.
They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.
Output
For each case, print the case number first. After that print "Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, thenprint the result where M is as small as possible.
Sample Input | Output for Sample Input |
3 10 5 12 | Case 1: 5 2 Case 2: Impossible Case 3: 3 4 |
w为偶数时 1*w==w所以必有解,遍历一下
代码:
/* 注意 long long */ #include<cstdio> int main() { int t,kcase=1; long long w,i; scanf("%d",&t); while(t--) { scanf("%lld",&w); if(w&1) { printf("Case %d: Impossible\n",kcase++); continue; } else { for(i=2;i<=w;i++) { if(w%i==0&&(w/i)&1) { printf("Case %d: %lld %lld\n",kcase++,w/i,i); break; } } } } return 0; }
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