Codeforces Round #699 (Div. 2)

Codeforces Round #699 (Div. 2)

A. K-divisible Sum

You are given two integers n and k.
You should create an array of n positive integers a1,a2,…,an such that the sum (a1+a2+⋯+an) is divisible by k and maximum element in a is minimum possible.
What is the minimum possible maximum element in a?
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.
The first and only line of each test case contains two integers n and k (1≤n≤109; 1≤k≤109).
Output
For each test case, print one integer — the minimum possible maximum element in array a such that the sum (a1+⋯+an) is divisible by k.

代码：

int T;
int main(){
    sd1(T);
    while(T--){
        LL n, k;
        sd2(n, k);
        LL ans = 0;
        if(k >= n){
            if(k % n == 0) ans = k / n;
            else  ans = k / n + 1;
        }
        else{
            if(n % k == 0){
                ans = 1;
            }
            else{
                ans = 2;
            }
        }
        cout << ans << endl;
    }
    system("pause");
    return 0;
}

B. Inflation

You have a statistic of price changes for one product represented as an array of n positive integers p0,p1,…,pn−1, where p0 is the initial price of the product and pi is how the price was increased during the i-th month.
Using these price changes you are asked to calculate the inflation coefficients for each month as the ratio of current price increase pi to the price at the start of this month (p0+p1+⋯+pi−1).
Your boss said you clearly that the inflation coefficients must not exceed k %, so you decided to increase some values pi in such a way, that all pi remain integers and the inflation coefficients for each month don’t exceed k %.
You know, that the bigger changes — the more obvious cheating. That’s why you need to minimize the total sum of changes.
What’s the minimum total sum of changes you need to make all inflation coefficients not more than k %?
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains two integers n and k (2≤n≤100; 1≤k≤100) — the length of array p and coefficient k.
The second line of each test case contains n integers p0,p1,…,pn−1 (1≤pi≤109) — the array p.
Output
For each test case, print the minimum total sum of changes you need to make all inflation coefficients not more than k %.

思路：从前向后贪心即可，每次使要加的数加到第一个上
代码：

int T;
LL a[MAXN];
LL sum[MAXN];
int main(){
    sd1(T);
    while(T--){
        ms(sum, 0);
        int n, k;
        sd2(n, k);
        LL summ = 0;
        for(int i = 1; i <= n; i++){
            sl1(a[i]);
            summ += a[i];
        }
        sum[1] = 0;
        LL ans = 0;
        for(int i = 2; i <= n; i++){
            sum[i] = sum[i-1] + a[i-1];
            if( (double)a[i]/(double)sum[i] * 100 > k ){
                sum[i] = ceil((double)a[i] / k * 100);
            }
        }
        cout << sum[n] - summ + a[n] << endl;
    }
    system("pause");
    return 0;
}

C. Longest Simple Cycle

You have n chains, the i-th chain consists of ci vertices. Vertices in each chain are numbered independently from 1 to ci along the chain. In other words, the i-th chain is the undirected graph with ci vertices and (ci−1) edges connecting the j-th and the (j+1)-th vertices for each 1≤j<ci.
Now you decided to unite chains in one graph in the following way:
the first chain is skipped;
the 1-st vertex of the i-th chain is connected by an edge with the ai-th vertex of the (i−1)-th chain;
the last (ci-th) vertex of the i-th chain is connected by an edge with the bi-th vertex of the (i−1)-th chain.
Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains the single integer n (2≤n≤105) — the number of chains you have.
The second line of each test case contains n integers c1,c2,…,cn (2≤ci≤109) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a1,a2,…,an (a1=−1; 1≤ai≤ci−1).
The fourth line of each test case contains n integers b1,b2,…,bn (b1=−1; 1≤bi≤ci−1).
Both a1 and b1 are equal to −1, they aren’t used in graph building and given just for index consistency. It’s guaranteed that the sum of n over all test cases doesn’t exceed 105.
Output
For each test case, print the length of the longest simple cycle.

思路：
从后向前贪心即可，注意一下加到第一个的时候就行
代码：

int T;
int c[MAXN], a[MAXN], b[MAXN];
int main(){
    sd1(T);
    while(T--){
        int n;
        sd1(n);
        int sum_n = 0;
        int sum = 0;
        for(int i = 0; i < n; i++){ sd1(c[i]);}
        for(int i = 0; i < n; i++){ sd1(a[i]);}
        for(int i = 0; i < n; i++){ sd1(b[i]);}
        sum_n = c[n-1];
        for(int i = n-1; i >= 1; i--){
            if(a[i] != b[i]){
                if(i == 1) {
                    sum_n += abs(a[i] - b[i]) + 1; 
                    sum = max(sum_n, sum);
                }
                else{
                    sum_n += (c[i-1] - abs(a[i] - b[i])+1);
                    sum_n = max(sum_n, c[i-1]);
                }
            }
            else{
                sum_n += 1;
                sum = max(sum_n, sum);
                if(i != 1) 
                    sum_n = c[i-1];
            }
        }
        cout << sum << endl;
    }
    system("pause");
    return 0;
}