Codeforces Round #514 (Div. 2), problem: (A) Cashier
阿新 • • 發佈:2018-12-13
給定客人來的時間和要接待他們的時間,然後算出自己能夠有多少次的休息
挨個列舉每一段的空閒時間就好啦,第i段的空閒時間是 t[i] - t[i-1] - last[i-1],然後除以a求和即可。
程式碼如下:
#include <bits/stdc++.h> using namespace std; typedef long long LL; int gcd(int a,int b){if (b == 0) return a; return gcd(b , a%b);} int lcm(int a, int b){ return a/gcd(a,b)*b;} inline int read(){ int f = 1, x = 0;char ch = getchar(); while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();} while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();} return x * f; } const int maxn = 1e5 + 10; int t[maxn],last[maxn]; int main(){ int n = read(),l = read(),a = read(); for (int i=1; i<=n; i++) { t[i] = read(); last[i] = read(); } int ans = 0; for (int i=1; i<=n; i++) { ans += (t[i] - last[i-1] - t[i-1])/a; } ans += (l - t[n] - last[n])/a; cout << ans << endl; return 0; }