More than 2019 cattle off summer school camp (seventh)

Question number	title	It has adopted the code	Interpretations / Discussion	Passing rate	Team status
A	String	Click to view	Enter discussions	561/3518	by
B	Irreducible Polynomial	Click to view	Enter discussions	724/2269	by
C	Governing sand	Click to view	Enter discussions	383/2052	by
D	Number	Click to view	Enter discussions	955/1462	by
E	Find the median	Click to view	Enter discussions	82/968	Did not pass
F	Energy stones	Click to view	Enter discussions	15/131	Did not pass
G	Make Shan Happy	Click to view	Enter discussions	2/28	Did not pass
H	Pair	Click to view	Enter discussions	92/239	Did not pass
I	Chessboard	Click to view	Enter discussions	17/82	Did not pass
J	A+B problem	Click to view	Enter discussions	1023/1843	by
K	Function	Click to view	Enter discussions	8/49	Did not pass

A

Knowledge Point: Minimum string representation

I to the beginning position of the string, looking from the back violence furthest j such that S [i ... j] represents the minimum of its own

ps: solution to a problem which did not engage in the practice of writing out, saw a lot of heavyweights on over less than 5ms

About smallest representation: OI Wiki

#include <bits/stdc++.h>
using namespace std;
int T,n;
int num;
char s[220];
char t[500];
//判断s[l...r] 是否为自己的最小表示
bool ok(int l,int r){
    int n = r - l + 1;
    for(int i=1;i<=n;i++)t[i] = s[l + i - 1];
    for(int i=1;i<=n;i++)t[n+i] = s[l + i - 1];
    int i=1,j=2,k;
    while(i<=n && j <= n){
        for(k = 0; k <= n && t[i+k] == t[j+k];k++);
            if(k == n)break;
        if(t[i+k] > t[j+k]){
            i = i + k + 1;
            if(i == j)i++;
        }
        else{
            j = j + k + 1;
            if(i == j)j++;
        }
    }
    int ans = min(i,j);
    for(int i=0;i<n;i++){
        if(t[ans+i] == s[l+i])continue;
        else return false;
    }
    return true;
}
void print(int l,int r){
    for(int i=l;i<=r;i++)printf("%c",s[i]);
    printf(" ");
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%s",s+1);
        n = strlen(s+1);
        num = 0;
        for(int i=1;i<=n;){
            for(int j=n;j>=i;j--){
                if(i == j || ok(i,j)){
                    print(i,j);
                    i = j+1;
                }
            }
        }
        puts("");
    }
    return 0;
}

B

Conclusion: real domain can not be split only two polynomials: a quadratic polynomial and ( \ (2 ^ B <4ac \) )

#include <bits/stdc++.h>
using namespace std;
int T;
int n;
long long a[1005];
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 0; i <= n; i++){
            scanf("%lld", &a[i]);
        }
        if(n >= 3){
            puts("No");
        }
        else if(n <= 1){
            puts("Yes");
        }
        else{
            if(a[1] * a[1] - 4 * a[0] * a[2] < 0){
                puts("Yes");
            }
            else{
                puts("No");
            }
        }
    }
    return 0;
}

C

According to the height of small to large order, and then enumerate the current height as the maximum height, then than it is high tree to be cut down, less than it was inside the tree to pick some number of trees (from small to large by price considerations) so that the remaining quantity is less than equal to the number of the current enumeration of the maximum height of the tree.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll cost[205];
 
struct node {
    int h, c, p;
}E[100005];
 
bool cmp(node A, node B) {
    return A.h > B.h;
}
ll suf[100005];
 
int main() {
    int n;
    while(~scanf("%d", &n)) {
        ll ans = 0;
        memset(cost, 0, sizeof(cost));
        for(int i = 1; i <= n; i++) {
            scanf("%d%d%d", &E[i].h, &E[i].c, &E[i].p);
            cost[E[i].c] += 1LL * E[i].p;
            ans += 1LL * E[i].c * E[i].p;
        }
        sort(E + 1, E + 1 + n, cmp);
        suf[n] = 1LL * E[n].p;
        for(int i = n - 1; i >= 1; i--) suf[i] = suf[i + 1] + 1LL * E[i].p;
 
        ll pre = 0;
        for(int i = 1; i <= n; i++) {
            int j;
            ll num = 0;    //同一高度的树
            ll cutx = 0;   //砍掉同一高度的花费
            bool tail = true;
            ll tmp = suf[i];
            for(j = i; j <= n; j++) {
                if(E[j].h == E[i].h) {
                    cutx += 1LL * E[j].p * E[j].c;
                    num += 1LL * E[j].p;
                    tmp -= 1LL * E[j].p;
                    cost[E[j].c] -= 1LL * E[j].p;
                } else {
                    tail = false;
                    i = j - 1;
                    break;
                }
            }
            if(tail) i = n;
 
            if(tmp < num) tmp = 0;
            else tmp = tmp - num + 1;
            ll cut = 0;
            for(int j = 1; j <= 200; j++) {
                if(cost[j] >= tmp) {
                    cut += 1LL * j * tmp;
                    tmp = 0;
                    break;
                } else {
                    cut += 1LL * j * cost[j];
                    tmp -= cost[j];
                }
            }
            ll ccc = cut + pre;
            ans = min(ans, ccc);
            pre += cutx;
            if(pre >= ans) break;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

D

If n is greater than or equal length p, the first output p 0 then back up p, otherwise the answer does not exist

#include <bits/stdc++.h>
using namespace std;
int n,p;
int calc(int x){
    int res = 0;
    while(x)res++,x/=10;
    return res;
}
int main(){
    scanf("%d%d",&n,&p);
    int len = calc(p);
    if(n >= len){
        printf("%d",p);
        for(int i=len;i<n;i++)printf("0");
        puts("");return 0;
    }
    else puts("T_T");
    return 0;
}

J

#include <bits/stdc++.h>
 
using namespace std;
 
int T;
 
char a[100];
char b[100];
 
char c[105];
 
int main(){
    scanf("%d", &T);
    while(T--){
        memset(c, 0, sizeof(c));
 
        scanf("%s%s", a, b);
        reverse(a, a + strlen(a));
        reverse(b, b + strlen(b));
        long long aa, bb;
        aa = bb = 0;
        for(int i = 0; a[i]; i++){
            aa *= 10;
            aa += a[i] - '0';
        }
        for(int i = 0; b[i]; i++){
            bb *= 10;
            bb += b[i] - '0';
        }
        long long ans = aa + bb;
         
        int cnt = 0;
        bool flag = false;
        for(int i = 0; ans > 0; i++){
            c[i] = '0' + (ans % 10);
             
            if(!flag && c[i] != '0'){
                flag = true;
                cnt = i;
            }
 
            ans /= 10;
        }
 
        if(!flag){
            puts("0");
        }
        else{
            printf("%s\n", c + cnt);
        }
    }
    return 0;
}