A - Ice Tea Store
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
Problem Statement
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
- 1≤Q,H,S,D≤108
- 1≤N≤109
- All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Sample Input 1
20 30 70 90 3
Sample Output 1
150
Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90+30+30=150 yen.
Sample Input 2
10000 1000 100 10 1
Sample Output 2
100
Even though a 2-liter bottle costs just 10 yen, you need only 1 liter. Thus, you have to buy a 1-liter bottle for 100 yen.
Sample Input 3
10 100 1000 10000 1
Sample Output 3
40
Now it's better to buy four 0.25-liter bottles for 10+10+10+10=40 yen.
Sample Input 4
12345678 87654321 12345678 87654321 123456789
Sample Output 4
1524157763907942
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 1044266558 #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; ll a[5],n; int main() { while(scanf("%lld%lld%lld%lld",&a[0],&a[1],&a[2],&a[3])!=EOF) { scanf("%lld",&n); for(int i=1;i<4;i++) a[i]=min(a[i],2*a[i-1]); printf("%lld ",n/2*a[3]+(n&1)*a[2]); } return 0; }
B - Reverse and Compare
Time limit : 2sec / Memory limit : 256MB
Score : 500 points
Problem Statement
You have a string A=A1A2…An consisting of lowercase English letters.
You can choose any two indices i and j such that 1≤i≤j≤n and reverse substring AiAi+1…Aj.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
- 1≤|A|≤200,000
- A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Sample Input 1
aatt
Sample Output 1
5
You can obtain aatt (don't do anything), atat (reverse A[2..3]), atta (reverse A[2..4]), ttaa (reverse A[1..4]) and taat (reverse A[1..3]).
Sample Input 2
xxxxxxxxxx
Sample Output 2
1
Whatever substring you reverse, you'll always get xxxxxxxxxx.
Sample Input 3
abracadabra
Sample Output 3
44
预处理。
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 1044266558 #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; ll a[200005][30],ans; char s[200005]; int main() { while(scanf("%s",s+1)!=EOF) { memset(a,0,sizeof(a)); int len=strlen(s+1); ans=0; for(int i=1;i<=len;i++) { for(int j=0;j<26;j++) a[i][j]=a[i-1][j]; a[i][s[i]-'a']++; } for(int i=1;i<=len;i++) ans+=len-i-a[len][s[i]-'a']+a[i][s[i]-'a']; printf("%lld ",ans+1); } return 0; }