Gym 100531B Buffcraft (贪心+暴力+前缀和)

题意：给定两个加血的方式，一个是直接加多少，另一种是加百分之几，然后你能够你选 k 种，问你选哪 k 种。

析：首先肯定要选加的多的，所以我们先排序，从大到小，然后用前缀和存储一下，再去枚举从第一种和从第二种选 i 个，从另一个中选 k-i的，

注意这个 k 可能大于 m+n，讨论一下。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int num, id;
    bool operator < (const Node &p) const{
        return num > p.num;
    }
};
Node a[maxn], b[maxn];
LL sum1[maxn], sum2[maxn];

void print(int n, Node *a){
    for(int i = 1; i <= n; ++i)
        if(1 == i)  printf("%d", a[i].id);
        else printf(" %d", a[i].id);
    printf("\n");
}

int main(){
    freopen("buffcraft.in", "r", stdin);
    freopen("buffcraft.out", "w", stdout);
    int base, k;
    while(scanf("%d %d %d %d", &base, &k, &n, &m) == 4){
        for(int i = 1; i <= n; ++i){
            scanf("%d", &a[i].num);
            a[i].id = i;
        }
        for(int i = 1; i <= m; ++i){
            scanf("%d", &b[i].num);
            b[i].id = i;
        }
        if(k >= m + n){
            printf("%d %d\n", n, m);
            print(n, a);
            print(m, b);
            continue;
        }
        sort(a+1, a+n+1);
        sort(b+1, b+m+1);
        sum1[0] = sum2[0] = 0;
        for(int i = 1; i <= n; ++i) sum1[i] = sum1[i-1] + a[i].num;
        for(int i = 1; i <= m; ++i) sum2[i] = sum2[i-1] + b[i].num;
        LL ans = 0;
        int idx = 0;
        for(int i = 0; i <= n && i <= k; ++i){
            if(m < k - i) continue;
            LL tmp = (base + sum1[i]) * (100 + sum2[k-i]);
            if(tmp > ans){
                ans = tmp;
                idx = i;
            }
        }
        for(int i = 0; i <= m && i <= k; ++i){
            if(n < k - i)  continue;
            LL tmp = (base + sum1[k-i]) * (100 + sum2[i]);
            if(tmp > ans){
                ans = tmp;
                idx = -i;
            }
        }

        if(idx < 0)  idx += k;
        printf("%d %d\n", idx, k - idx);
        print(idx, a);
        print(k - idx, b);
    }
    return 0;
}