POJ 3388 Japanese Puzzle (二分)

题意：给你一个n*n 的图，你总共有k 种花砖，告诉你每一种花砖的个数，让你随便安排它们的位置，问你最多有多少行和第一行是一样，并且要输出第一行的一定存在的图案。

析：首先这个题如果读懂了题意，一点也不难，就是一个普通的二分，可是我真的是读不懂啊，尤其是这个输出解的时候，我以为是输出每行存在编号，真是被坑死了。现在分析怎么二分，就是直接二分答案，假设是 mid，然后对于每种图案有 val 个，要想每行都有，那么在每行中最多就有 val / mid 次，最后检查一下，这个值的和是不是大于等于 n 就OK了。

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>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#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 fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "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 = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 5e4 + 5;
const int maxm = 700 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -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;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

struct Node{
  int id, x;
  bool operator < (const Node &p) const{
    return x > p.x;
  }
};
Node a[maxn];

bool judge(int mid){
  int ans = 0;
  for(int i = 0; i < m && a[i].x / mid; ++i)
    ans += a[i].x / mid;
  return ans >= n;
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 0; i < m; ++i){
      scanf("%d", &a[i].x);
      a[i].id = i;
    }
    sort(a, a + m);
    int l = 1, r = n;
    while(l <= r){
      int mid = l + r >> 1;
      if(judge(mid))  l = mid + 1;
      else r = mid - 1;
    }
    printf("%d\n", r);
    int sum = 0;
    for(int i = 0; i < m; ++i){
      int t = a[i].x / r;
      if(sum + t < n)  FOR(j, t, 0)   printf("%d\n", a[i].id+1);
      else{
        FOR(j, n-sum, 0)  printf("%d\n", a[i].id + 1);
        break;
      }
      sum += t;
    }
  }
  return 0;
}