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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#100546#5419. Triangleslgvc#WA 2ms3604kbC++231.9kb2023-04-26 17:56:482023-04-26 17:56:52

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-04-26 17:56:52]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:3604kb
  • [2023-04-26 17:56:48]
  • 提交

answer

//这回只花了114514min就打完了。
//真好。记得多手造几组。
#include <bits/stdc++.h>
//#define int long long
#define pai 3.141592653589793238462643383279502884197169399375105820
#define MOD 1000000007
#define eps 0.00000001
inline int min(int a,int b) {return a<b?a:b;}
inline int max(int a,int b) {return a>b?a:b;}
#define ULL unsigned long long
#define LL long long
#define INF 0x3f3f3f3f
#define INF_LL 0x3f3f3f3f3f3f3f3f
static char buf[1000000],*paa=buf,*pd=buf;
static char buf2[1000000],*pp=buf2;
#define getchar() paa==pd&&(pd=(paa=buf)+fread(buf,1,1000000,stdin),paa==pd)?EOF:*paa++
inline void pc(char ch){
	if(pp-buf2==1000000) fwrite(buf2,1,1000000,stdout),pp=buf2;
	*pp++=ch;
}
inline void pcc(){
	fwrite(buf2,1,pp-buf2,stdout);
	pp=buf2;
}
inline int read(void) {
	int w=1;
	register int x(0);register char c(getchar());
	while(c<'0'||c>'9') {if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return w*x;
}
void write(int x) {
	if(x<0) pc('-'),x=-x;
	static int sta[20];
	int top=0;
	do {
		sta[top++]=x%10,x/=10;
	} while(x);
	while(top) pc(sta[--top]+48);
}
void we(int x) {
	write(x);
	pc('\n');
}
inline bool cmp_xi(int a,int b) {return a<b;}
inline bool cmp_da(int a,int b) {return a>b;}
int N;
signed main(void) {
    //freopen("m.in","r",stdin);
    //freopen("m.out","w",stdout);
	scanf("%d",&N);
	if(N<=3) {
		printf("No");
		return 0;
	}
//	printf("0 0 0 1000000000 1000000000 0\n");
//	printf("1000000000 1000000000 0 1000000000 1000000000 0\n");
	printf("Yes\n");
	printf("0 0 0 1000000000 1 999999999\n");
	printf("0 0 1000000000 0 1 999999999\n");
	for(int i=1;i<N-2;i++) {
		printf("1000000000 1000000000 %d %d %d %d\n",i-1,1000000000-i+1,i,1000000000-i);
	}
	printf("1000000000 1000000000 %d %d 1000000000 0\n",N-3,1000000000-(N-3));
    return 0;
}


Details

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Test #1:

score: 100
Accepted
time: 2ms
memory: 3604kb

input:

2

output:

No

result:

ok no solution

Test #2:

score: -100
Wrong Answer
time: 2ms
memory: 3484kb

input:

24

output:

Yes
0 0 0 1000000000 1 999999999
0 0 1000000000 0 1 999999999
1000000000 1000000000 0 1000000000 1 999999999
1000000000 1000000000 1 999999999 2 999999998
1000000000 1000000000 2 999999998 3 999999997
1000000000 1000000000 3 999999997 4 999999996
1000000000 1000000000 4 999999996 5 999999995
1000000...

result:

wrong answer triangle 1 not acute