HDU 5881 Tea (模拟)

题意：有一壶水, 体积在 LLL 和 RRR 之间, 有两个杯子, 你要把水倒到两个杯子里面, 使得杯子水体积几乎相同(体积的差值小于等于1),

并且使得壶里剩下水体积不大于1. 你无法测量壶里剩下水的体积, 问最小需要倒水的次数。

析：考虑倒水的大致过程，不妨设 L > 0。首先向一个杯子倒 L/2 ?2??L?? 升水，再往另一个杯子倒 L/2+1?2??L??+1 升水。接下来就来回往两个杯子里倒 2 升，

直到倒空为止。这样就很容易分析出需要倒水的次数。唯一注意的是最后壶里面可以剩下 1 升水，可以省一次倒水的操作。

就是一步一步的模拟就可以，注意特殊情况，并且这个特殊情况较多。

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>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"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 int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    LL l, r;
    while(scanf("%I64d %I64d", &l, &r) == 2){
        if(!l){
            if(r <= 1)  printf("0\n");
            else printf("%I64d\n", (r+1)/2);
        }
        else if(1 == l){
            if(r == 1)  printf("0\n");
            else printf("%I64d\n", (1+r)/2);
        }
        else if(2 == l){
            if(r == 2)  printf("1\n");
            else if(r >= 3 && r <= 5)  printf("2\n");
            else printf("%I64d\n", (r-l-2)/2+2);
        }
        else{
            LL ans = 2;
            ans += (r-l-2)/2;
            ans = Max(ans, 2LL);
            printf("%I64d\n", ans);
        }
    }
    return 0;
}