poj 3273 二分答案

原題連結

​​http://poj.org/problem?id=3273​​

題意

思路

參考代碼

#include <algorithm>
#include <cmath>
#include <cstdio>

using namespace std;
int n, m, a[100005];
bool check(int mid) {
  int sum = 0, num = 1; //查詢每組上限為mid的情況下的組數
  for (int i = 1; i <= n; i++) {
    if (sum + a[i] <= mid) {
      sum += a[i];
    } else {
      sum = a[i];
      num++;
    }
  }
  return num > m; //組數偏大，說明mid偏小
}
int main() {
  scanf("%d%d", &n, &m);
  int maxn = -1, sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maxn = max(maxn, a[i]);
    sum += a[i];
  }
  int l = maxn, r = sum;
  int mid = (l + r) / 2;
  while (l < r) {
    if (check(mid)) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
    mid = (l + r) / 2;
  }
  printf("%d\n", mid);
  return 0;
}