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	444 KiB
#3	Accepted 5	1ms	444 KiB
#4	Accepted 5	1ms	404 KiB
#5	Accepted 5	2ms	552 KiB
#6	Accepted 5	1ms	404 KiB
#7	Accepted 5	11ms	1.8 MiB
#8	Accepted 5	10ms	1.6 MiB
#9	Accepted 5	1ms	404 KiB
#10	Accepted 5	1ms	404 KiB
#11	Accepted 5	12ms	1.6 MiB
#12	Accepted 5	12ms	1.8 MiB
#13	Accepted 5	1ms	404 KiB
#14	Accepted 5	2ms	404 KiB
#15	Accepted 5	13ms	1.6 MiB
#16	Accepted 5	12ms	1.5 MiB
#17	Accepted 5	2ms	404 KiB
#18	Accepted 5	1ms	404 KiB
#19	Accepted 5	12ms	1.6 MiB
#20	Accepted 5	12ms	1.7 MiB

#include<bits/stdc++.h>
using namespace std;
int t,n,a[100005][2],a2[100005][2],top2,b2[100005],top,b[100005],ans;
string s1,s2,t1,t2;
int main(){
    freopen("edit.in","r",stdin);
    freopen("edit.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        cin>>s1>>s2>>t1>>t2;
        top=1;
        a[top][0]=0,a[top][1]=0;
        top2=1;
        a2[top2][0]=0,a2[top2][1]=0;
        ans=0;
        for(int i=0;i<n;i++){
            if(t1[i]=='1'){
                if(s1[i]=='0')a[top][0]++;
                else a[top][1]++;
                b[i]=top;
            }else{
                top++;
                a[top][0]=0;
                a[top][1]=0;
                b[i]=0;
            }
            if(t2[i]=='1'){
                if(s2[i]=='0')a2[top2][0]++;
                else a2[top2][1]++;
                b2[i]=top2;
            }else{
                top2++;
                a2[top2][0]=0;
                a2[top2][1]=0;
                b2[i]=0;
            }
        }
        for(int i=0;i<n;i++){
            if(t1[i]=='1'&&t2[i]=='1'){
                if(a[b[i]][0]>0&&a2[b2[i]][0]>0){
                    a[b[i]][0]--,a2[b2[i]][0]--;
                    ans++;
                }else if(a[b[i]][1]>0&&a2[b2[i]][1]>0){
                    a[b[i]][1]--,a2[b2[i]][1]--;
                    ans++;
                }
            }else if(t1[i]=='0'&&t2[i]=='1'){
                if(s1[i]=='1'&&a2[b2[i]][1]>0){
                    a2[b2[i]][1]--;
                    ans++;
                }else if(s1[i]=='0'&&a2[b2[i]][0]>0){
                    a2[b2[i]][0]--;
                    ans++;
                }
            }else if(t1[i]=='1'&&t2[i]=='0'){
                if(s2[i]=='1'&&a[b[i]][1]>0){
                    a[b[i]][1]--;
                    ans++;
                }else if(s2[i]=='0'&&a[b[i]][0]>0){
                    a[b[i]][0]--;
                    ans++;
                }
            }else{
                if(s1[i]==s2[i])ans++;
            }
        }
        cout<<ans<<"\n";
    }
    return 0;
}

递交者: 248
题目: NOIP2024A NOIP2024A
比赛: 【民间数据】NOIP2024 批量测试TT
语言: C++14(O2)
代码长度: 2.1 KiB
递交时间: 2024-12-1 11:55:33
评测时间: 2024-12-1 12:11:27

分数: 100
总耗时: 111ms
峰值时间: 13ms
峰值内存: 1.8 MiB

100 Accepted

代码

信息

100 Accepted

代码

信息

还没有账户？

登录