Codeforces #371(Div.2)A.Meeting of Old Friends【模拟+思维】
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Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minutek she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.
Calculate the number of minutes they will be able to spend together.
The only line of the input contains integers l1,r1,l2, r2 andk (1 ≤ l1, r1, l2, r2, k ≤ 1018,l1 ≤ r1,l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
1 10 9 20 1
2
1 100 50 200 75
50
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
题目大意:
有两个人,一个人会呆从l1到r1 这么长时间,另一个人会呆从l2,r2这么长时间,但是第一个人会在K这个时间点去化妆(一分钟),问 两个人同时在一起的时间为多少。
思路:
1、首先取l1,l2中大的,作为相同区间的左边界,然后取r1,r2中小的,作为相同区间的右边界,然后最后判一下k是否在区间内。
2、注意如果没有相同区间,千万不要输出负值。
Ac代码:
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define ll __int64int main(){ ll l1,r1,l2,r2,k; while(~scanf("%I64d%I64d%I64d%I64d%I64d",&l1,&r1,&l2,&r2,&k)) { ll output=0; if(l1<=l2) { ll you=min(r1,r2); output+=you-l2+1; if(k>=l2&&k<=you)output--; } else { ll you=min(r1,r2); output+=you-l1+1; if(k>=l1&&k<=you)output--; } if(output<0)output=0; printf("%I64d\n",output); fflush(stdout); }}
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