POJ 3273 - Monthly Expense(二分搜索)

题目:

http://poj.org/problem?id=3273

题意:

将n(1 ≤ N ≤ 100,000)天分成m个月, 给出n天的花费mon(1 ≤ moneyi ≤ 10,000),求出m个月中最大花费的最小值.

思路:

二分搜索. 左值l = 0, 右值r= MAZ_N*MAX_MONEY, (这个wa了一次).

CODE:

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int INF = 1e9+5;
int mon[100005], n, m;

bool cor(int x)
{
int tmp = 0;
for(int i = 0; i < m; ++i) {
int s = 0, cnt = tmp - 1;
while(cnt < n && s < x) {
cnt++;
s += mon[cnt];
}
if(cnt == n) return false;
//printf("%d\n", s - mon[cnt]);
tmp = cnt;
}
return true;
}
void solve()
{
int l = 0, r = INF;
while(r - l > 1) {
int mid = (r + l) / 2;
if(cor(mid)) l = mid;
else r = mid;
}
printf("%d\n", l);
}
int main()
{
//freopen("in", "r", stdin);
while(~scanf("%d %d", &n, &m)) {
memset(mon, 0, sizeof(mon));
for(int i = 0; i < n; ++i) {
scanf("%d", &mon[i]);
}
solve();
}
return 0;
}