2019 Multi-University Training Contest 4

Solved	Pro.ID	Title	Ratio(Accepted / Submitted)
	1001	AND Minimum Spanning Tree	31.75%(1018/3206)
	1002	Colored Tree	0.00%(0/105)
	1003	Divide the Stones	10.35%(126/1217)
	1004	Enveloping Convex	0.00%(0/125)
	1005	Good Numbers	17.65%(9/51)
	1006	Horse	29.03%(18/62)
	1007	Just an Old Puzzle	31.83%(875/2749)
	1008	K-th Closest Distance	10.97%(376/3429)
	1009	Linear Functions	0.00%(0/35)
Have been added	1010	Minimal Power of Prime	6.15%(328/5331)

1001

1 and the even-even point, a first for the odd dot positions to find the binary representation of 0, so that the bit is a binary 1, and if greater than n, then the odd and one pair, or pairs with the number of

#include <bits/stdc++.h>
using namespace std;
int ans[200005];
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        int anss = 0;
        for(int i = 2; i <= n; i++) {
            if(i % 2 == 0) ans[i] = 1;
            else {
                int tmp = i;
                int cnt = 0;
                bool f = false;
                while(tmp) {
                    if(tmp & 1) cnt++;
                    else {
                        f = true;
                        break;
                    }
                    tmp >>= 1;
                }
                if(f) ans[i] = 1 << cnt;
                else {
                    if((1 << cnt) <= n) ans[i] = 1 << cnt;
                    else ans[i] = 1, anss++;
                }
            }
        }
        printf("%d\n", anss);
        for(int i = 2; i <= n; i++) {
            if(i != n) printf("%d ", ans[i]);
            else printf("%d\n", ans[i]);
        }
    }
    return 0;
}

1007

See the results in the initial state are the same reverse sequence of the parity relationship between the position difference and the difference between the number of row 0 is located.

#include <bits/stdc++.h>
using namespace std;
int a[7][7];
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int ans = 0;
        for(int i = 1; i <= 4; i++) {
            for(int j = 1; j <= 4; j++) {
                scanf("%d", &a[i][j]);
                if(a[i][j] == 0) ans += 4 - i;
            }
        }
        ans %= 2;
        int res = 0;
        for(int i = 1; i <= 4; i++) {
            for(int j = 1; j <= 4; j++) {
                if(a[i][j] == 0)continue;
                for(int k1 = i; k1 <= 4; k1++) {
                    int t;
                    if(k1 == i) t = j + 1;
                    else t = 1;
                    for(int k2 = t; k2 <= 4; k2++) {
                        if(a[k1][k2] == 0) continue;
                        if(a[i][j] > a[k1][k2]) res++;
                    }
                }
            }
        }
        res %= 2;
        if((res == 1 && ans == 1)||(res == 0 && ans == 0)){
            puts("Yes");
        }
        else{
            puts("No");
        }
    }
    return 0;
}

1010

x maximum 1e18, observe the decomposition of the quality factor, you can traverse relatively small prime factors, then the rest will not be too great, for example, prime numbers near 1000, index of 6 to 1e18, so when we put less than 1200 after (or 1100) of prime numbers all taken into account, the rest of that number may be \ (the p-^ 1 \) , \ (the p-^ 2 \) , \ (p_1 ^ 2p_2 ^ 2 \) , \ (the p-^ 3 \ ) , \ (. 4 ^ P \) , \ (^. 5 P \) , \ (P_1 3p_2 ^ 2 ^ \) found no good judge smallest prime. So we put this case extended to the case of index 5, the \ (x ^ {1 \ over 5} \) prime factors decomposed, and the rest could only be \ (P ^. 1 \) , \ ( 2 ^ P \) , \ (P ^. 3 \) , \ (2p_2 ^ P ^ 2 \) , \ (. 4 ^ P \)

So just do a square root calculation then determine whether the same power like

Standard A:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1500;
int p[N],v[N],tot;
void getP(){
    for(int i=2;i<N;i++){
        if(!v[i])p[++tot] = i;
        for(int j=1;j<=tot && p[j] * i < N;j++){
            v[p[j] * i] = 1;
            if(i % p[j] == 0)break;
        }
    }
}
inline ll po(ll a,int b){
  ll ret=1;
  while(b--)ret*=a;
  return ret;
}
inline bool ok(ll x,int k){
    int t = round(pow(x,1.0/k));
    if(po(t,k) == x)return true;
    else return false;
}
inline int solve(ll x){
    int res = 100;
    for(int i=1;i<=tot && p[i] <= x ;i++){
        int z = 0;
        while(x % p[i] == 0){
            x /= p[i];
            z ++;
        }
        if(z)res = min(res,z);
    }
    if(x == 1)return res;
    for(int i=5;i>=2;i--){
        if(ok(x,i)){
            return res = min(res,i);
        }
    }
    return 1;
}
int main(){
    getP();
    int T;scanf("%d",&T);
    ll x;
    while(T--){
        scanf("%lld",&x);
        printf("%d\n",solve(x));
    }
    return 0;
}