Codeforces 460D Little Victor and Set (constructed), codeforces460d

Codeforces 460D Little Victor and Set (constructed), codeforces460d

Link: Codeforces 460D Little Victor and Set

For a given range of l and r, select a number smaller than k to minimize or equal to these numbers.

Solution: If k is an even number, then kXOR (k + 1) = 1
Based on this, all the situations where k is less than 3 can be processed.
For k = 3, find a 2 k greater than l. If the value is 2 k + 1 ≤ r, you can create three numbers or the value is 0.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll l, r, k;void solve (int n) {    int s = 1;    ll ret = l;    for (int i = 2; i < (1<<n); i++) {        ll u = 0;        for (int j = 0; j < n; j++)            if (i&(1<<j))                u ^= (l+j);        if (u < ret) {            ret = u;            s = i;        }    }    int x = 0;    ll arr[10];    for (int i = 0; i < n; i++)        if (s&(1<<i))            arr[x++] = l + i;    printf("%lld\n%d\n", ret, x);    printf("%lld", arr[0]);    for (int i = 1; i < x; i++)        printf(" %lld", arr[i]);    printf("\n");}void fuck () {    ll x = 1;    while (x <= l) x <<= 1;    if ((x | (x>>1)) <= r) {        printf("0\n3\n");        printf("%lld %lld %lld\n", l, x | (x>>1), (x | (x>>1)) ^ l);        return;    }    printf("1\n2\n");    if (l&1)        l++;    printf("%lld %lld\n", l, l+1);}int main () {    scanf("%lld%lld%lld", &l, &r, &k);    if (k == 1)        printf("%lld\n1\n%lld\n", l, l);    else if (k >= 5) {        if (l&1)            l++;        printf("0\n4\n%lld %lld %lld %lld\n", l, l+1, l+2, l+3);    } else if (k == 4 && (r - l + 1 > 4 || l%2 == 0)) {        if (l&1)            l++;        printf("0\n4\n%lld %lld %lld %lld\n", l, l+1, l+2, l+3);    } else if (k == 2 && (r - l + 1 > 2 || l%2 == 0)) {        if (l&1)            l++;        printf("1\n2\n%lld %lld\n", l, l+1);    } else if (k == 2) {        if ((l^r) > l)            printf("%lld\n1\n%lld\n", l, l);        else            printf("%lld\n2\n%lld %lld\n", l^r, l, r);    } else if (k == 4) {        solve(k);    } else {        fuck();    }    return 0;}