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	448 KiB
#3	Accepted 5	1ms	444 KiB
#4	Accepted 5	1ms	404 KiB
#5	Accepted 5	2ms	448 KiB
#6	Accepted 5	1ms	404 KiB
#7	Accepted 5	16ms	2.9 MiB
#8	Accepted 5	15ms	3 MiB
#9	Accepted 5	1ms	404 KiB
#10	Accepted 5	1ms	512 KiB
#11	Accepted 5	15ms	3 MiB
#12	Accepted 5	16ms	3 MiB
#13	Accepted 5	1ms	408 KiB
#14	Accepted 5	1ms	452 KiB
#15	Accepted 5	16ms	2.9 MiB
#16	Accepted 5	15ms	2.9 MiB
#17	Accepted 5	1ms	404 KiB
#18	Accepted 5	1ms	404 KiB
#19	Accepted 5	15ms	3 MiB
#20	Accepted 5	14ms	3 MiB

#include <bits/stdc++.h>
using namespace std;
int n,nxt1[100005],nxt2[100005],cnt1[100005][2],cnt2[100005][2];
char s1[100005],s2[100005],t1[100005],t2[100005];
int main(){
    freopen("edit.in","r",stdin);
    freopen("edit.out","w",stdout);
    int TTTQ;
    cin>>TTTQ;
    while(TTTQ--){
        scanf("%d%s%s%s%s",&n,s1+1,s2+1,t1+1,t2+1);
        nxt1[n+1]=nxt2[n+1]=n+1;
        cnt1[n+1][0]=cnt1[n+1][1]=cnt2[n+1][0]=cnt2[n+1][1]=0;
        for(int i=n;i>=1;i--){
            cnt1[i][0]=cnt1[i][1]=cnt2[i][0]=cnt2[i][1]=0;
            if(t1[i]=='0'){
                nxt1[i]=i;
            } else{
                nxt1[i]=nxt1[i+1];
            }
            if(t2[i]=='0'){
                nxt2[i]=i;
            } else{
                nxt2[i]=nxt2[i+1];
            }
        }
        for(int i=1;i<=n;i++){
            if(t1[i]=='1')
                cnt1[nxt1[i]-1][s1[i]-48]++;
            if(t2[i]=='1')
                cnt2[nxt2[i]-1][s2[i]-48]++;
        }
        int ans=0;
        for(int i=1;i<=n;i++){
            bool b10,b11,b20,b21;
            if(t1[i]=='1'){
                b10=(cnt1[nxt1[i]-1][0]>0);
                b11=(cnt1[nxt1[i]-1][1]>0);
            } else{
                b10=(s1[i]=='0');
                b11=(s1[i]=='1');
            }
            if(t2[i]=='1'){
                b20=(cnt2[nxt2[i]-1][0]>0);
                b21=(cnt2[nxt2[i]-1][1]>0);
            } else{
                b20=(s2[i]=='0');
                b21=(s2[i]=='1');
            }
            if(b10&&b20){
                if(t1[i]=='1'){
                    cnt1[nxt1[i]-1][0]--;
                }
                if(t2[i]=='1'){
                    cnt2[nxt2[i]-1][0]--;
                }
                ans++;
            } else if(b11&&b21){
                if(t1[i]=='1'){
                    cnt1[nxt1[i]-1][1]--;
                }
                if(t2[i]=='1'){
                    cnt2[nxt2[i]-1][1]--;
                }
                ans++;
            }
        }
        printf("%d\n",ans);
    }

    return 0;
}

递交者: LN-0142
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 KiB
递交时间: 2024-12-1 15:26:48
评测时间: 2024-12-1 15:38:15

分数: 100
总耗时: 139ms
峰值时间: 16ms
峰值内存: 3 MiB

100 Accepted

代码

信息

100 Accepted

代码

信息

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