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. 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)
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 2Sample Output
Yes No
题意:判断能否组成四边形。
quadrilateral | |
英[ˌkwɒdrɪˈlætərəl] | 美[ˌkwɑ:drɪˈlætərəl] |
adj. |
n. |
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; }