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HDU 5881 Tea (模拟)

2024-08-13 17:46:53   221人阅读

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;}

 

HDU 5881 Tea (模拟)


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