CodeForces 670D2 Magic Powder

题意：今天我们要来造房子。造这个房子需要n种原料，每造一个房子需要第i种原料a_i个。现在你有第i种原料b_i个。此外，你还有一种特殊的原料k个，

每个特殊原料可以当作任意一个其它原料使用。那么问题来了，你最多可以造多少个房子呢？

析：首先可以先把开始能造出的先处理出来，然后再进行二分，当然也可以直接进行二分。

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>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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 bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn];

bool judge(LL mid){
  LL mm = m;
  for(int i = 0; i < n; ++i){
    if(b[i] / a[i] >= mid)  continue;
    mm += b[i] - a[i] * mid;
    if(mm < 0)  return false;
  }
  return true;
}

int main(){
  scanf("%d %d", &n, &m);
  LL sum = 0;
  for(int i = 0; i < n; ++i){
    scanf("%d", a+i);
    sum += a[i];
  }
  int ans = INF;
  for(int i = 0; i < n; ++i){
    scanf("%d", b+i);
    ans = min(ans, b[i] / a[i]);
  }
  for(int i = 0; i < n; ++i)  b[i] -= a[i] * ans;

  LL l = 0, r = m;
  while(l < r){
    int mid = l + (r-l+1) / 2;
    if(judge(mid))  l = mid;
    else r = mid - 1;
  }
  cout << ans + l << endl;
  return 0;
}