HDU - 5655 CA Loves Stick(竟有四边形判定定理。。。)
2017-08-03 14:56
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CA loves to play with sticks.
One day he receives four pieces of sticks, he wants to know these sticks can spell a quadrilateral.
(What is quadrilateral? Click here: https://en.wikipedia.org/wiki/Quadrilateral)
InputFirst line contains TT denoting
the number of testcases.
TT testcases
follow. Each testcase contains four integers a,b,c,da,b,c,d in
a line, denoting the length of sticks.
1≤T≤1000, 0≤a,b,c,d≤ 263−11≤T≤1000, 0≤a,b,c,d≤263−1
OutputFor each testcase, if these sticks can spell a quadrilateral, output "Yes"; otherwise, output "No" (without the quotation marks).
Sample Input
Sample Output
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5655
题意:判断能否组成四边形。
[例句]Isoparametric element of arbitrary quadrilateral plays
an important role in finite element analysis.
等参数任意四边形单元在有限元分析中有着重要的作用。
一:不知道这个单词的意思,不会看懂题的。
二:知道这个题什么意思,不一定知道有这么一个定理,判断四边形。
(a<=b<=c<=d) d<a+b+c;
三:知道前两点也要让你错几次!!!。数据太坑了。。2^63-1,把人往死里坑
。
我用的是d-b-c<a.稍不注意就会数据崩溃。
还要注意0!!!
One day he receives four pieces of sticks, he wants to know these sticks can spell a quadrilateral.
(What is quadrilateral? Click here: https://en.wikipedia.org/wiki/Quadrilateral)
InputFirst line contains TT denoting
the number of testcases.
TT testcases
follow. Each testcase contains four integers a,b,c,da,b,c,d in
a line, denoting the length of sticks.
1≤T≤1000, 0≤a,b,c,d≤ 263−11≤T≤1000, 0≤a,b,c,d≤263−1
OutputFor each testcase, if these sticks can spell a quadrilateral, output "Yes"; otherwise, output "No" (without the quotation marks).
Sample Input
2 1 1 1 1 1 1 9 2
Sample Output
Yes No
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5655
题意:判断能否组成四边形。
quadrilateral | |
英[ˌkwɒdrɪˈlætərəl] | 美[ˌkwɑ:drɪˈlætərəl] |
adj. | 四边(形)的; |
n. | 四边形; |
an important role in finite element analysis.
等参数任意四边形单元在有限元分析中有着重要的作用。
一:不知道这个单词的意思,不会看懂题的。
二:知道这个题什么意思,不一定知道有这么一个定理,判断四边形。
(a<=b<=c<=d) d<a+b+c;
三:知道前两点也要让你错几次!!!。数据太坑了。。2^63-1,把人往死里坑
。
我用的是d-b-c<a.稍不注意就会数据崩溃。
还要注意0!!!
#include<stdio.h> #include<algorithm> using namespace std; __int64 a[5]; int main() { int t; scanf("%d",&t); while(t--) { int f=0; for(int i=0; i<4; i++) { scanf("%I64d",&a[i]); if(a[i]==0)f=1; } if(f) { printf("No\n"); continue; } sort(a,a+4); if(a[3]-a[2]-a[1]<a[0]) { printf("Yes\n"); } else printf("No\n"); } return 0; }
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