Codeforces Round #326 (Div. 2)

Codeforces Round #326 (Div. 2)


A. Duff and Meat

水題就不多說了 一直記錄最小值就可以了

__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 只要用快速mod的方式去找出每個質因數，而如果答案沒辦法整除這個質因數就把這個質因數乘起來

__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 只要有兩個以上的A數字 就可以mod 2變成只有1個or 0個 只剩1個的話代表無法合併 為一個step 而map(A+1)可以加上A / 2的數量 並且一直往大的數字推

__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);
}

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