2014
02-27

# Warcraft

Have you ever played the Warcraft?It doesn’t matter whether you have played it !We will give you such an experience.There are so many Heroes in it,but you could only choose one of them.Each Hero has his own skills.When such a Skill is used ,it costs some MagicValue,but hurts the Boss at the same time.Using the skills needs intellegence,one should hurt the enemy to the most when using certain MagicValue.

Now we send you to complete such a duty to kill the Boss(So cool~~).To simplify the problem:you can assume the LifeValue of the monster is 100, your LifeValue is 100,but you have also a 100 MagicValue!You can choose to use the ordinary Attack(which doesn’t cost MagicValue),or a certain skill(in condition that you own this skill and the MagicValue you have at that time is no less than the skill costs),there is no free lunch so that you should pay certain MagicValue after you use one skill!But we are good enough to offer you a "ResumingCirclet"(with which you can resume the MagicValue each seconds),But you can’t own more than 100 MagicValue and resuming MagicValue is always after you attack.The Boss is cruel , be careful!

There are several test cases,intergers n ,t and q (0<n<=100，1<=t<=5，q>0) in the first line which mean you own n kinds of skills ,and the "ResumingCirclet" helps you resume t points of MagicValue per second and q is of course the hurt points of LifeValue the Boss attack you each time(we assume when fighting in a second the attack you show is before the Boss).Then n lines follow,each has 2 intergers ai and bi(0<ai,bi<=100).which means using i skill costs you ai MagicValue and costs the Boss bi LifeValue.The last case is n=t=q=0.

There are several test cases,intergers n ,t and q (0<n<=100，1<=t<=5，q>0) in the first line which mean you own n kinds of skills ,and the "ResumingCirclet" helps you resume t points of MagicValue per second and q is of course the hurt points of LifeValue the Boss attack you each time(we assume when fighting in a second the attack you show is before the Boss).Then n lines follow,each has 2 intergers ai and bi(0<ai,bi<=100).which means using i skill costs you ai MagicValue and costs the Boss bi LifeValue.The last case is n=t=q=0.

4 2 25
10 5
20 10
30 28
76 70
4 2 25
10 5
20 10
30 28
77 70
0 0 0

4
My god

Hint
Hint:
When fighting,you can only choose one kind of skill or just to use the  ordinary attack in the whole second,the ordinary attack costs the Boss 1
points of LifeValue,the Boss can only use ordinary attack which costs a whole second at a time.Good Luck To You!


dp[i][j+b[k]] = max( dp[i][j+b[k]],
dp[i-1][j]-a[i]
);

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iomanip>
using namespace std;

int dp[105][105];///dp[i][j]表示攻击了i次，打了对方j点生命，自己剩下的魔值
int a[105]= {0};
int b[105]= {1};

int main() {
int n,t,k;
while(cin>>n>>t>>k,n+t+k) {
int m=100/k+(100%k?1:0);
for(int i=1; i<=n; ++i)
cin>>a[i]>>b[i];
memset(dp,-1,sizeof(dp));
dp[0][0]=100;///开始时魔法值为100
for(int i=1;i<=m;++i)
for(int j=0;j<=100;++j)
if(dp[i-1][j]!=-1){
dp[i-1][j] = (dp[i-1][j]+t)>100 ? 100:dp[i-1][j]+t;///魔值不超过100
for(int k=0;k<=n;++k)
if(dp[i-1][j]>=a[k]&&dp[i-1][j]-a[k]>dp[i][j+b[k]])
if(j+b[k]>=100)
dp[i][100]=dp[i-1][j]-a[k];
else
dp[i][j+b[k]]=dp[i-1][j]-a[k];
}
int p=1;
for( ;p<=m;++p)
if(dp[p][100]!=-1){
cout<<p<<endl; break;
}
if(p>m)puts("My god");
}
return 0;
}

1. #!/usr/bin/env python
def cou(n):
arr =
i = 1
while(i<n):
arr.append(arr[i-1]+selfcount(i))
i+=1
return arr[n-1]

def selfcount(n):
count = 0
while(n):
if n%10 == 1:
count += 1
n /= 10
return count

2. I go through some of your put up and I uncovered a good deal of expertise from it. Many thanks for posting this sort of exciting posts

3. 代码是给出了，但是解析的也太不清晰了吧！如 13 abejkcfghid jkebfghicda
第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)，为什么要这样拆分，原则是什么？