2025钉耙编程-中国大学生算法设计暑期联赛（8）——最努力的活着 · znzryb
## 思路讲解
暴力程序是这样的。
```cpp
i128 ans=0;
while(true){
if(n<W) break;
ans+=(1+N)*N/2-(sum+1)*sum/2;
sum+=n/W;
n-=n/W;
}
ans+=(1+N)*N/2-(sum+1)*sum/2;
```
不难发现，我们可以分块处理相同n/W的部分，以获得logn的时间复杂度。
## AC代码
[https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14726](https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14726)
::: 源代码
```cpp
// by Gospel_rock
#include <bits/stdc++.h>
#include <cstdio>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#define fsp(x) fixed<<setprecision(x)
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
#endif
using ll = __int128;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
i128 N,M,Q,X,K,T,lT,W;
inline void __();
namespace QuickRead { // 快读
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T> void Cin(T &a) {
T ans = 0;
bool f = 0;
char c = getc();
for (; c < '0' || c > '9'; c = getc()) {
if (c == '-') f = 1;
}
for (; c >= '0' && c <= '9'; c = getc()) {
ans = ans * 10 + c - '0';
}
a = f ? -ans : ans;
}
template <typename T, typename... Args> void Cin(T &a, Args &...args) {
Cin(a), Cin(args...);
}
template <typename T> void write(T x) { // 注意，这里输出不带换行
if (x < 0) putchar('-'), x = -x;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
} // namespace QuickRead
using namespace QuickRead;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
Cin(lT);
while(lT--)
__();
return 0;
}
/*
*/
i128 ans=0;
// void calres(ll n,ll obj,ll step){
// // i128 l=(i128)N*(N-1)/2-(i128)(N-obj+1)*(N-obj)/2;
// // i128 r=(i128)N*(N-1)/2-(i128)(N-n+1)*(N-n)/2;
// i128 num=n-obj;
// i128 l=(i128)(N-n+1)*(N-n)/2;
// i128 r=(i128)(N-obj+1)*(N-obj)/2;
// i128 res=(i128)(N+1)*N/2*((n-obj+1)/step)-(l+r)*((n-obj+1)/step)/2;
// #ifdef LOCAL
// debug(n);debug(obj);
// debug((ll)l);debug((ll)r);debug((ll)res);
// cend;
// #endif
// ans+=res;
// }
void calres(i128 n,i128 obj,i128 step,i128 num){
i128 sum=(N+1)*N/2*num;
i128 fn=N-n,fob=N-(obj+step);
// =(fn+jd)^2
i128 sumsq=num*(num-1)*(2*num-1)/6;
i128 minus=fn*fn*num+sumsq*step*step+2*fn*(num-1+0)*num/2*step;
minus+=(fn+fob)*num/2;
minus/=2;
ans+=sum-minus;
}
// ceil_div
template<typename T> T chu(T x,T y) {
if ((x ^ y) < 0) return x / y;
if (x % y == 0) return x / y;
return x / y + 1;
}
inline void __(){
Cin(N,W);
i128 n=N;
ans=0;
while(true){
if(n<W) break;
i128 step=n/W;
i128 obj=W*(n/W)-1;
ll num=chu((n-obj),step);
obj=n-step*num;
calres(n, obj,n/W,num);
n=obj;
}
ans+=(1+N)*N/2-(i128)(N-n+1)*(N-n)/2;
write(ans);putchar('\n');
}
/*
前面因为long long溢出WA了一发
AC
https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14726
(1+N)*N/2-(sum+1)*sum/2:15
(1+N)*N/2-(sum+1)*sum/2:12
(1+N)*N/2-(sum+1)*sum/2:9
(1+N)*N/2-(sum+1)*sum/2:5
1
7 4
n:7
(1+N)*N/2-(sum+1)*sum/2:28
n:6
(1+N)*N/2-(sum+1)*sum/2:27
n:5
(1+N)*N/2-(sum+1)*sum/2:25
n:4
(1+N)*N/2-(sum+1)*sum/2:22
n:3
(1+N)*N/2-(sum+1)*sum/2:18
*/
```
:::
## 心路历程（WA，TLE，MLE……）
容易发现，fob的计算一定是交给后面算的，就算是最后一个也是。
