Atcoder contest104 ABC C - All Green 暴力 加 贪心

C - All Green

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Time limit : 2sec / Memory limit : 1024MB

Score : 300 points

Problem Statement

A programming competition site AtCode provides algorithmic problems. Each problem is allocated a score based on its difficulty. Currently, for each integer i between 1 and D (inclusive), there are pi problems with a score of 100i points. These p1+…+pD problems are all of the problems available on AtCode.

A user of AtCode has a value called total score. The total score of a user is the sum of the following two elements:

• Base score: the sum of the scores of all problems solved by the user.
• Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of ci points, aside from the base score (1≤i≤D).

Takahashi, who is the new user of AtCode, has not solved any problem. His objective is to have a total score of G or more points. At least how many problems does he need to solve for this objective?

Constraints

• 1≤D≤10
• 1≤pi≤100
• 100≤ci≤106
• 100≤G
• All values in input are integers.
• ci and G are all multiples of 100.
• It is possible to have a total score of G or more points.

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Input

Input is given from Standard Input in the following format:

D G
p1 c1
:
pD cD

Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

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Sample Input 1

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2 700
3 500
5 800

Sample Output 1

Copy

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

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Sample Input 2

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2 2000
3 500
5 800

Sample Output 2

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7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

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Sample Input 3

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2 400
3 500
5 800

Sample Output 3

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2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

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Sample Input 4

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5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

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66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.、

 

题意:给出你d个题目，然后和一个需要达到的分数G，对于每个题目，第i个题目做一道会获得 i * 100的分数，如果

做到p[i]的数目的话，就可以获得额外的分数c[i]。每道题目，只能做一套

 

思路: 因为最多只有10个题目，那我们可以先暴力每个题目做与不做，就是2^10，如果有剩下的分数，你就可以用

贪心，倒着取题目，来做，然后对于每次的题目数取个min

 

 

代码如下：

#include <bits/stdc++.h>
 
using namespace std;
 
 
const int maxn = 200;
int d,gg,c[maxn],p[maxn],vis[maxn];
struct node {
    int c,p;
    int id,s;
} g[maxn];
 
int getans() {
    int ans = 0;
    int sum = 0;
    int G = gg;
    for(int i = 1; i <= d; i++) {
        if(vis[i]) G -= (g[i].p * 100 * i + g[i].c),ans += g[i].p;
    }
    for(int i = d; i >= 1; i--) {
        if(!vis[i]) {
            if(G > 0) {
                if(G >= g[i].p * 100 * i) {
                    ans += g[i].p;
                    G -= (g[i].p * 100 * i + g[i].c);
 
                } else {
                    ans += G / (i * 100);
                    G -= 100 * i * (G / (i * 100));
                }
            }
 
        }
    }
    if(G <= 0) return ans;
    else return 1e9;
}
int main() {
    while(cin >> d >> gg) {
        int ans = 1e9;
        for(int i = 1; i <= d; i++)
            cin >> g[i].p >> g[i].c,g[i].id = i;
        for(int i = 0; i < (1 << d); i++) {
            for(int j = 1; j <= d; j++) {
                if(1 & (i >> (j - 1))) vis[j] = 1;
                else vis[j] = 0;
            }
            int sum = getans();
            ans = min(ans,sum);
        }
        cout << ans << endl;
    }
    return 0;
}