• Codeforces 985 D


    D - Sand Fortress

    思路:

    二分

    有以下两种构造,

    分别二分取个最小。

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    #define fi first
    #define se second
    #define pi acos(-1.0) 
    #define LL long long 
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pii pair<int, int> 
    #define mem(a, b) memset(a, b, sizeof(a))
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
    //head
    ULL n, h;
    bool check(ULL t) {
        ULL res;
        if(t >= h) res = (1+t)*t/2 + t*(t-1)/2 - h*(h-1)/2;
        else res = (1+t)*t/2;
        if(res >= n) return true;
        else return false;
    } 
    bool Check(ULL t) {
        ULL res;
        if(t >= h) res = (1+t)*t - h*(h-1)/2;
        else res = (1+t)*t/2;
        if(res >= n) return true;
        else return false;
    }
    int main() {
        scanf("%llu%llu", &n, &h);
        ULL l = 0, r = 2e9, m = (l+r) >> 1;
        while(l < r) {
            if(check(m)) r = m;
            else l = m+1;
            m = (l+r) >> 1;
        }
        ULL ans;
        if(m >= h) ans = 2*m - h;
        else ans = m;
        l = 0, r = 2e9, m = (l+r) >> 1;
        while(l < r) {
            if(Check(m)) r = m;
            else l = m+1;
            m = (l+r) >> 1;
        }
        if(m >= h) ans = min(ans, 2*m - h + 1);
        else ans = min(ans, m);
        printf("%llu
    ", ans); 
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9074529.html
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