A monster is attacking the Cyberland!

Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attributes: hitpoints (HP), offensive power (ATK) and defensive power (DEF).

During the battle, every second the monster‘s HP decrease by max(0, ATKY - DEFM), while Yang‘s HP decreases bymax(0, ATKM - DEFY), where index Y denotes Master Yang and index M denotes monster. Both decreases happen simultaneously Once monster‘s HP ≤ 0 and the same time Master Yang‘s HP > 0, Master Yang wins.

Master Yang can buy attributes from the magic shop of Cyberland: h bitcoins per HP, a bitcoins per ATK, and d bitcoins per DEF.

Now Master Yang wants to know the minimum number of bitcoins he can spend in order to win.

The first line contains three integers HPY, ATKY, DEFY, separated by a space, denoting the initial HP, ATK and DEF of Master Yang.

The second line contains three integers HPM, ATKM, DEFM, separated by a space, denoting the HP, ATK and DEF of the monster.

The third line contains three integers h, a, d, separated by a space, denoting the price of 1 HP, 1 ATK and 1 DEF.

All numbers in input are integer and lie between 1 and 100 inclusively.

The only output line should contain an integer, denoting the minimum bitcoins Master Yang should spend in order to win.

暴力攻防

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;const int maxa = 300;int dp[maxa][maxa];int main(){    int x, y, z;    int x1, y1, z1;    int a, b, c;    cin>>x>>y>>z>>x1>>y1>>z1>>a>>b>>c;    int guanwujianxue = y -  z1;    int uu = 0;             //钱    if(guanwujianxue <= 0){        uu = b * (-guanwujianxue + 1);        guanwujianxue = 1;    }    int yingxiongjianxue = max(0, y1 - z);    int mina = 10000000;    for(int i =guanwujianxue; i < maxa; i++){        for(int k= yingxiongjianxue; k >= 0; k--){            int sum = (i - guanwujianxue)*b + (yingxiongjianxue-k)*c;            int n = x1/i;            if(x1 % i != 0)n++;            if(k * n < x)                mina = min(mina, sum);            else{                mina = min(mina, sum + (k*n+1-x)*a);            }        }    }    cout<<mina+uu<<endl;}

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Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

思路就是线性的，看到个牛逼的解法

#include<stdio.h>#include<string.h>#include<iostream>#include<set>using namespace std;const int maxa = 100005;int dp[maxa];int n, s, l;multiset<int>st, rt;int a[maxa];int main(){    scanf("%d%d%d", &n, &s, &l);    for(int i = 0; i < n; i++){        scanf("%d", &a[i]);    }    for(int i = 0, j = 0; i < n; i++){        st.insert(a[i]);        while(*st.rbegin() - *st.begin() > s){            st.erase(st.find(a[j]));            if(i - j >= l)                rt.erase(rt.find(dp[j-1]));            j++;        }        if(i - j+1 >=l)rt.insert(dp[i-l]);        if(rt.begin() == rt.end())dp[i] = maxa;            else dp[i] = *rt.begin()+1;    }    if(dp[n-1] >= maxa)dp[n-1] = -1;    cout<<dp[n-1]<<endl;}

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