题意：
谷仓里最多有n 个谷子， 每天早上都会运来m 个 ，多了的就不要了，第i 天呢会有i 只鸟来吃谷子， 每一个鸟都会吃，都只吃一个，问你哪一天谷仓为空？
思路：
分类讨论了：
当n <= m时候 ans = n
当n > m时候 我们发现 前m 天 一直都是满的 ，不可能为空， 第m+1天早上 满的，晚上 会少m-1个， 第m+2天早上又运来m 个，下午没了m-2个，
那么假设第x 天没了的话，
那么 m +1 - m + m + 2 -m + m +3 - m ....+ m + x >= n
即1 + 2 +3 +。。。 + x + m >= n
即 ((1+x)*x)/2 + m >= n 二分答案即可。
别忘了加上前面的m 天。
#include <bits/stdc++.h>
#define mr make_pair
#define fi first
#define se second
#define ps push_back
#define Siz(x) (int)x.size()
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-10;
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 200000 + 10;
LL n,m;
bool judge(LL x){
return (x*x+x)/2+m >= n;
}
int main(){
scanf("%lld %lld",&n, &m);
if (n <= m) return 0 * printf("%lld\n",n);
LL L = 1, R = (LL)2e9+1;
LL ans;
while(L <= R){
LL m = L+R>>1LL;
if (judge(m)){
ans = m;
R = m-1;
}
else L = m+1;
}
printf("%lld\n",ans + m);
return 0;
}
C. Anton and Fairy Tale
time limit per test
memory limit per test
input
output
Anton likes to listen to fairy tales, especially when Danik, Anton's best friend, tells them. Right now Danik tells Anton a fairy tale:
"Once upon a time, there lived an emperor. He was very rich and had much grain. One day he ordered to build a huge barn to put there all his grain. Best builders were building that barn for three days and three nights. But they overlooked and there remained a little hole in the barn, from which every day sparrows came through. Here flew a sparrow, took a grain and flew away..."
n
- m grains are brought to the barn. If m
- i-th day i
Anton is tired of listening how Danik describes every sparrow that eats grain from the barn. Anton doesn't know when the fairy tale ends, so he asked you to determine, by the end of which day the barn will become empty for the first time. Help Anton and write a program that will determine the number of that day!
Input
n and m (1 ≤ n, m ≤ 1018)
Output
Output one integer — the number of the day when the barn will become empty for the first time. Days are numbered starting with one.
Examples
input
5 2
output
4
input
8 1
output
5
Note
In the first sample the capacity of the barn is five grains and two grains are brought every day. The following happens:
- At the beginning of the first day grain is brought to the barn. It's full, so nothing happens.
- 5 - 1 = 4
- At the beginning of the second day two grains are brought. The barn becomes full and one grain doesn't fit to it.
- 5 - 2 = 3
- At the beginning of the third day two grains are brought. The barn becomes full again.
- 5 - 3 = 2
- 2 + 2 = 4
- 4 - 4 = 0
4, because by the end of the fourth day the barn becomes empty.
