Codeforces Round #326 (Div. 2)

Oct 18, 2015#CodeForce #math59

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This is a Codeforces competition round with the title "Codeforces Round #326 (Div. 2)" held on October 18, 2015. It includes a problem called "Duff and Meat" where the task is to record the minimum value. The code provided calculates the solution for the problem.

Codeforces Round #326 (Div. 2)

A. Duff and Meat

There is no need to say much about this easy problem, just keep track of the minimum value.

\_\_author\_\_ = 'GCA'
# Date: 2015/10/18

n = int(input())
minp = 200
total = 0
for i in range(n):
    ai, pi = list(map(int, input().split()))
    if pi < minp:
        minp = pi
    total += minp \* ai
print(total)

B. Duff in Love Just use the fast mod method to find each prime factor, and if the answer cannot be divided by this prime factor, multiply it.

\_\_author\_\_ = 'GCA'
# Date: 2015/10/19
n = int(input())
a = n \*\* 0.5 + 1
d = 2
A = 1
while d <= a:
    if n % d == 0:
        n //= d
        if A % d != 0:
            A \*= d
    else:
        d += 1
if n > 1:
    A \*= n
print(A)


```C. Duff and Weight Lifting As long as there are more than two A numbers, mod 2 can be used to become only 1 or 0. If there is only 1 left, it means that it cannot be merged and becomes one step. And map(A+1) can add the quantity of A / 2, and keep pushing to larger numbers.

__author__ = 'GCA'
// Created by GCA on 2015/10/19

#include using namespace std;

const int maxn = 1000105;
int a[maxn];
int n;
int main() {
memset(a, 0, sizeof(a));
scanf("%d", &n);
for (int i = 0; i < n; i++) {
int t;
scanf("%d", &t);
a[t]++;
}
int ans=0;
for (int i = 0; i < maxn; i++) {
ans += a[i]%2;
a[i+1]+=a[i]/2;
}
printf("%d\n",ans);
}