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QOJ
| ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
|---|---|---|---|---|---|---|---|---|---|
| #861187 | #9982. Staircase Museum | ucup-team4717# | WA | 5ms | 8008kb | C++14 | 684b | 2025-01-18 16:26:08 | 2025-01-18 16:29:48 |
Judging History
answer
#include<bits/stdc++.h>
#define int long long
#define N 500005
using namespace std;
int T;
int n;
int l[N],r[N];
int f[N][2];
void sol(){
scanf("%lld",&n);
for(int i=1;i<=n;i++) scanf("%lld %lld",&l[i],&r[i]);
for(int i=1;i<=n;i++){
f[i][0]=f[i-1][0];
f[i][1]=f[i-1][1];
if(l[i]!=l[i-1]) f[i][0]=f[i-1][0]+1;
if(r[i]!=r[i-1]) f[i][1]=f[i-1][1]+1;
if(i>=3&&r[i-2]<l[i]) f[i][0]=max(f[i][0],f[i-2][1]+2);
if(i>=3&&l[i-2]<l[i-1]&&r[i-1]<r[i]) f[i][1]=max(f[i][1],f[i-2][0]+2);
}
printf("%lld\n",max(f[n][0],f[n][1]));
}
signed main(){
scanf("%lld",&T);
while(T--) sol();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 8008kb
input:
4 3 1 2 1 3 1 3 3 1 2 2 3 3 3 3 1 1 1 3 3 3 4 1 2 2 3 3 4 4 5
output:
2 3 3 4
result:
ok 4 number(s): "2 3 3 4"
Test #2:
score: 0
Accepted
time: 0ms
memory: 8008kb
input:
1 1 1 1000000000
output:
1
result:
ok 1 number(s): "1"
Test #3:
score: -100
Wrong Answer
time: 5ms
memory: 7996kb
input:
9653 1 1 1 2 1 1 1 1 3 1 1 1 1 1 1 4 1 1 1 1 1 1 1 1 5 1 1 1 1 1 1 1 1 1 1 6 1 1 1 1 1 1 1 1 1 1 1 1 6 1 2 1 2 1 2 1 2 1 2 2 2 6 1 1 1 1 1 1 1 1 1 1 1 2 6 1 2 1 2 1 2 1 2 1 2 2 3 5 1 2 1 2 1 2 1 2 2 2 6 1 2 1 2 1 2 1 2 2 2 2 2 6 1 3 1 3 1 3 1 3 2 3 3 3 6 1 2 1 2 1 2 1 2 2 2 2 3 6 1 3 1 3 1 3 1 3 2 3...
output:
1 1 1 1 1 1 2 2 2 2 2 3 3 3 2 2 2 3 3 3 3 2 2 3 3 3 3 3 2 2 2 3 2 3 3 3 4 4 4 3 3 3 4 4 3 4 3 3 4 4 4 4 4 2 2 2 2 3 3 2 2 3 3 2 2 3 3 3 3 3 3 3 3 3 3 4 4 3 3 4 4 4 4 3 3 3 4 4 3 3 4 4 4 4 3 3 4 3 4 3 3 4 4 4 4 4 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 3 3 3 4 4 3 3 4 4 4 4 3 3 4 4 4 4 4 4 4 4 4 3 3 4 ...
result:
wrong answer 20th numbers differ - expected: '2', found: '3'