NOI批量评测 - 记录详情

#	状态分数	耗时	内存占用

#1	Accepted 5	1ms	404 KiB
#2	Accepted 5	1ms	316 KiB
#3	Accepted 5	2ms	408 KiB
#4	Accepted 5	1ms	408 KiB
#5	Accepted 5	1ms	404 KiB
#6	Accepted 5	2ms	404 KiB
#7	Accepted 5	9ms	1.2 MiB
#8	Accepted 5	9ms	1.1 MiB
#9	Accepted 5	2ms	448 KiB
#10	Accepted 5	2ms	404 KiB
#11	Accepted 5	8ms	1.2 MiB
#12	Accepted 5	8ms	1 MiB
#13	Accepted 5	2ms	404 KiB
#14	Accepted 5	2ms	404 KiB
#15	Accepted 5	8ms	1.2 MiB
#16	Accepted 5	8ms	1.3 MiB
#17	Accepted 5	2ms	404 KiB
#18	Accepted 5	1ms	452 KiB
#19	Accepted 5	10ms	1.2 MiB
#20	Accepted 5	10ms	1.2 MiB

#include <bits/stdc++.h>

using namespace std;

int T, n, cnt1[2], cnt2[2];
string s1, s2, t1, t2;

void Solve() {
    cin >> n >> s1 >> s2 >> t1 >> t2;
    s1 = ' ' + s1, s2 = ' ' + s2;
    t1 = ' ' + t1, t2 = ' ' + t2;

    int r1 = 1, r2 = 1, ans = 0;
    for (; r1 <= n && t1[r1] != '0'; r1++) cnt1[s1[r1] - '0']++;
    for (; r2 <= n && t2[r2] != '0'; r2++) cnt2[s2[r2] - '0']++;

    while (r1 <= n || r2 <= n) {
        if (r1 > r2) {
            int p = min(cnt1[0], cnt2[0]), q = min(cnt1[1], cnt2[1]);
            ans += p + q;
            cnt1[0] -= p + cnt2[1] - q, cnt1[1] -= q + cnt2[0] - p;
            cnt2[0] = cnt2[1] = 0;

            if (r2 <= n) {
                int t = s2[r2] - '0';
                if (cnt1[t]) ans++, cnt1[t]--;
                else cnt1[t ^ 1]--;
            }

            for (r2++; r2 <= n && t2[r2] != '0'; r2++) cnt2[s2[r2] - '0']++;
        } else if (r2 > r1) {
            int p = min(cnt1[0], cnt2[0]), q = min(cnt1[1], cnt2[1]);
            ans += p + q;
            cnt2[0] -= p + cnt1[1] - q, cnt2[1] -= q + cnt1[0] - p;
            cnt1[0] = cnt1[1] = 0;

            if (r1 <= n) {
                int t = s1[r1] - '0';
                if (cnt2[t]) ans++, cnt2[t]--;
                else cnt2[t ^ 1]--;
            }

            for (r1++; r1 <= n && t1[r1] != '0'; r1++) cnt1[s1[r1] - '0']++;
        } else {
            ans += min(cnt1[0], cnt2[0]) + min(cnt1[1], cnt2[1]) + (r1 <= n && s1[r1] == s2[r2]);
            cnt1[0] = cnt2[0] = cnt1[1] = cnt2[1] = 0;
            for (r1++; r1 <= n && t1[r1] != '0'; r1++) cnt1[s1[r1] - '0']++;
            for (r2++; r2 <= n && t2[r2] != '0'; r2++) cnt2[s2[r2] - '0']++;
        }
    }
    ans += min(cnt1[0], cnt2[0]) + min(cnt1[1], cnt2[1]);
    cout << ans << '\n';
    cnt1[0] = cnt2[0] = cnt1[1] = cnt2[1] = 0;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    freopen("edit.in", "r", stdin);
    freopen("edit.out", "w", stdout);
    cin >> T;
    while (T--) Solve();
    return 0;
}

递交者: HN-0285
题目: NOIP2024AGF NOIP2024AGF
比赛: 【官方数据 HN】 NOIP2024 批量测试
语言: C++14(O2)
代码长度: 2 KiB
递交时间: 2024-12-6 17:25:45
评测时间: 2024-12-6 17:27:38

分数: 100
总耗时: 87ms
峰值时间: 10ms
峰值内存: 1.3 MiB

100 Accepted

代码

信息

100 Accepted

代码

信息

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