NOI批量评测 - 记录详情

#	状态分数	耗时	内存占用

#1	Accepted 5	2ms	404 KiB
#2	Accepted 5	2ms	404 KiB
#3	Accepted 5	1ms	404 KiB
#4	Accepted 5	1ms	320 KiB
#5	Accepted 5	2ms	404 KiB
#6	Accepted 5	1ms	452 KiB
#7	Accepted 5	13ms	1.5 MiB
#8	Accepted 5	12ms	1.3 MiB
#9	Accepted 5	1ms	404 KiB
#10	Accepted 5	1ms	412 KiB
#11	Accepted 5	14ms	1.4 MiB
#12	Accepted 5	15ms	1.5 MiB
#13	Accepted 5	2ms	404 KiB
#14	Accepted 5	1ms	408 KiB
#15	Accepted 5	16ms	1.5 MiB
#16	Accepted 5	13ms	1.3 MiB
#17	Accepted 5	2ms	408 KiB
#18	Accepted 5	2ms	404 KiB
#19	Accepted 5	16ms	1.4 MiB
#20	Accepted 5	15ms	1.4 MiB

#include<bits/stdc++.h>
using namespace std;
#define up(s, e, i) for (int i = s; i <= e; i++)
#define down(s, e, i) for (int i = s; i >= e; i--)
#define read(s, e, a) for (int III = s; III <= e; III++) cin>>a[III]
#define all(x) x.begin(),x.end()
#define endl '\n'
namespace Main
{
    const int N = 100010;
    bool tk1[N], tk2[N];
    void solve()
    {
        int n;
        string s1, s2, t1, t2;
        cin>> n >> s1 >> s2 >> t1 >> t2;
        up(1, n, i) tk1[i] = tk2[i] = 0;
        s1 = " " + s1, s2 = " " + s2, t1 = " " + t1, t2 = " " + t2;
        int cnt10 = 0, cnt11 = 0, cnt20 = 0, cnt21 = 0, ans = 0;
        up(1, n, i)
        {
            if(t1[i]=='0')
            {
                cnt10 = cnt11 = 0;
                (s1[i] == '0' ? cnt10 : cnt11)++;
            }
            else
            {
                if(!tk1[i])
                {
                    up(i,n,j)
                    {
                        if(t1[j]=='0')
                            break;
                        tk1[j] = 1;
                        (s1[j] == '0' ? cnt10 : cnt11)++;
                    }
                }
            }
            
            if(t2[i]=='0')
            {
                cnt20 = cnt21 = 0;
                (s2[i] == '0' ? cnt20 : cnt21)++;
            }
            else
            {
                if(!tk2[i])
                {
                    up(i,n,j)
                    {
                        if(t2[j]=='0')
                            break;
                        tk2[j] = 1;
                        (s2[j] == '0' ? cnt20 : cnt21)++;
                    }
                }
            }
            if (cnt10 && cnt20)
                ans++, cnt10--, cnt20--;
            else if(cnt11&&cnt21)
                ans++, cnt11--, cnt21--;
            if(t1[i]=='0')
            {
                cnt10 = cnt11 = 0;
            }
            if(t2[i]=='0')
            {
                cnt20 = cnt21 = 0;
            }

        }
        cout << ans << endl;
    }
    void main()
    {
        freopen("edit.in", "r", stdin);
        freopen("edit.out", "w", stdout);
        ios::sync_with_stdio(0);
        cin.tie(0), cout.tie(0);
        int T;
        cin >> T;
        while(T--)
            solve();
    }
}
int main()
{
    Main::main();
    return 0;
}

// g++ edit.cpp -o edit -O2 -std=c++14 -Wall && ./edit

/*
1
5
11101
11110
11101
10111

1
5
00011
11000
01110
11011

*/

递交者: LN-0009
题目: NOIP2024A NOIP2024A
比赛: 【民间数据】NOIP2024 LN 批量测试
语言: C++14(O2)
代码长度: 2.4 KiB
递交时间: 2024-12-1 15:26:43
评测时间: 2024-12-1 15:33:20

分数: 100
总耗时: 131ms
峰值时间: 16ms
峰值内存: 1.5 MiB

100 Accepted

代码

信息

100 Accepted

代码

信息

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