典型的二分,二分上限是组成一个组,花费是每月花费之和,下线是组成N个组,花费是他们中最大的。单调性容易证明,分成组越少,花费越高,分成组越多,花费越少,呈线性关系,故可以采用二分法求解单调函数极值(最值)。
#include <iostream>
#include <vector>
using namespace std;
bool judge(int N, int M, int mid, vector <int> a) {
int sum = 0;
int group = 1;
for (int i = 0; i < N; i++) {
if (sum + a[i] <= mid) { // sum from day 0 to day i, if the sum <= mid, it means they must be in one group
sum += a[i];
}
else { // if sum from day 0 to day i-1, plus day i, is bigger than sum, then it should be divided in to two groups, day 1 to day i-1, day i.
sum = a[i];
group++;
}
}
if (group > M) {
return false;
}
else return true;
}
int main() {
int N, M;
vector <int> a;
while (cin >> N >> M) {
int max = 0, sum = 0;
for (int i = 0; i < N; i++) { //max group->n group->with max upon money, min group->1 group with sum of money
int temp;
cin >> temp;
a.push_back(temp);
if (temp > max) {
max = temp;
}
sum += temp;
}
int mid = (max + sum) >> 1;
while (max < sum) {
if (judge(N, M, mid, a)) {
sum = mid - 1;
}
else max = mid + 1;
mid = (sum + max) >> 1;
}
cout << mid << endl;
}
}