https://codeforces.com/contest/1562/problem/C

c

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <vector>
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
const int N=2e5+10,M=1e6+10;

int n,m;
int q[55],sum[10];
bool st[N];

void solve()
{
    int l0=0;
    cin>>n;
    string s;
    cin>>s;
    l0=s.find('0');
    if(l0==-1)
    {
        printf("1 %d 2 %d\n",n-1,n);
        return ;
    }
    else
    {
        if(l0+1<=n/2)
            printf("%d %d %d %d\n",l0+1,n,l0+2,n);
        else
            printf("%d %d %d %d\n",1,l0+1,1,l0);
    }
}

int main()
{
    int T;cin>>T;
    while(T--)
        solve();
    return 0;
}

D1

其实③也满足下面红字那句话 满足的条件 代码中归为一类

#include <iostream>

using namespace std;

int a[1000000 + 5], p[1000000 + 5];

int get_sum(int l, int r) {
    if (l > r) {
        return 0;
    }
    return (l % 2 == 1) ? p[r] - p[l - 1] : p[l - 1] - p[r];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin>>t;
    while (t--) {
        int n, q;
        cin >> n >> q;
        string ss;
        cin >> ss;
        for (int i = 1; i <= n; i++) {
            a[i] = (ss[i - 1] == '+' ? 1 : -1);
        }
        p[0] = 0;
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] + (i % 2 == 1 ? a[i] : -a[i]);
        }
        for (int o = 0; o < q; o++) {
            int l, r;
            cin >> l >> r;
            if (get_sum(l, r) == 0) {
                cout << "0\n";
                continue;
            }
            if ((r - l + 1) % 2 == 0) {
                cout << "2\n";
            }
            else {
                cout << "1\n";
            }
        }
    }
}