【CodeForces】Educational Codeforces Round 118 (Rated for Div. 2)【A-C】
A. Long Comparison
思路
- 先判断字符长度
- 相等的字符长度判断填满0后判断字典序
AC代码
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")#include<bits/stdc++.h>
using namespace std;typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;ll a[maxn];
ll b[maxn];int main(void){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endif int t;cin >> t;while(t--){string x1;int p1;string x2;int p2;cin >> x1 >> p1;cin >> x2 >> p2;int len1 = x1.size();int len2 = x2.size();if(len1+p1 > len2+p2) cout<<">"<<endl;else if(len1+p1 < len2+p2) cout<<"<"<<endl;else{if(len1==len2){if(x1 > x2) cout<<">"<<endl;else if(x1 < x2) cout<<"<"<<endl;else cout<<"="<<endl;}else{if(len1 < len2){for(int i = 1; i <= len2-len1; ++i) x1 += '0';if(x1>x2) cout<<">"<<endl;else if(x1<x2) cout<<"<"<<endl;else cout<<"="<<endl;}else{for(int i = 1; i <= len1-len2; ++i) x2 += '0';if(x1>x2) cout<<">"<<endl;else if(x1<x2) cout<<"<"<<endl;else cout<<"="<<endl;}}}}return 0;
}
B. Absent Remainder
思路
a%b<ba\%b < ba%b<b，找到最小的b即可。
AC代码
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")#include<bits/stdc++.h>
using namespace std;typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;ll a[maxn];
ll b[maxn];int main(void){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endif int t;cin >> t;while(t--){int n;cin >> n;for(int i = 1; i <= n; ++i) cin >> a[i];sort(a+1,a+1+n);for(int i = 1; i <= n/2; ++i){cout<<a[i+1]<<" "<<a[1]<<endl;}}return 0;
}
C. Poisoned Dagger
思路
二分K即可。
AC代码
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")#include<bits/stdc++.h>
using namespace std;typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;ll a[maxn];
ll b[maxn];
ll n,h;
ll check(ll mid){ll sum = 0;for(int i = 1; i <= n-1; ++i) sum += min(mid,a[i+1]-a[i]);sum += mid;return sum;
}
int main(void){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endif int t;cin >> t;while(t--){cin >> n >> h;for(int i = 1; i <= n; ++i) cin >> a[i];ll l = 1;ll r = h;while(l < r){ll mid = (l+r) >> 1;if(check(mid) < h) l = mid+1;else r = mid;}cout<<l<<endl;}return 0;
}
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