UVA - 11526 - H(n)（思路，减少枚举量）

Posted 2020-08-16 TianTengtt

题意：给定在32位带符号整数范围内的n，求n/1+n/2+n/3+n/4+……+n/n = ？

因为损失精度，所以算出来的有些连续的项是相同的数值，故想办法找出对于某个值，哪一段范围均是这个值。

详见代码注释

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;

#define NDEBUG
#include<cassert>
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;

int main(){
    ll n;
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%lld", &n);
        ll sum = 0;
        for(ll i = 1; i <= n;){
            ll value = n / i;       //此时需要找的值为value
            ll lur = n / value;     //n中最多包含lur个value，向下取整（因为不能是小数）
            sum += (lur - i + 1) * value;      // i ~ lur 范围内的值均是value
            i = lur + 1;            //减少枚举量
        }
        printf("%lld\n", sum);
    }
    return 0;
}