Problem Description
Little Rabbit loves playing a tabletop game called Monopoly. In the game, players roll the dice to move around the game board, buying and trading properties, and developing them with houses and hotels. Players collect rent from their opponents, with the goal being to drive them into bankruptcy.
One day, Little Rabbit invites Little Horse to play the game with him, but Little Horse stands him up. As a result, Little Rabbit can only play the game on his own. After a while, Little Rabbit finds out another way to play the game.
Let’s consider the map of the game as n integers a1,a2,…,an placed in a circle. If the player lands on ai, his score will change by ai points. Specifically, if ai>0, his score will increase by |ai| points. If ai<0, his score will decrease by |ai| points. And of course, if ai=0, his score won’t change.
As there is nobody else, Little Rabbit doesn’t feel like rolling a dice. So he just moves on the map step by step. Formally, if he now lands on ai and i<n, he will move to ai+1 in the next step. If he now lands on an, he will move to a1 in the next step.
Initially, Little Rabbit has 0 points, and his first step is to land on a1. Little Rabbit wonders how many steps he should take at least to reach exactly x points. (Specifically, since his initial score is 0, he needs 0 steps to reach 0 points.)
前话
看了不少题解好像都是要对模数为负数的时候进行特殊处理，但看不懂为啥，于是自己搞了一个不判断模数正负直接模的方法.不同语言 取模实现不一样，具体原理点这里
题解
设 a[i] 为前缀和
-
大体思路一样，搞了个map 套 map （第一个map，存同模数对应的同余域，第二个map存这个同余域里所有的{a[i],i } ） -
用a[i] % a[n]处理出的同余域，这里注意 如果 a[i] 和 a[n] 不同号，也要把{a[i],i} 加到 a[i] % a[n] + a[n] 的同余域 -
比如 在C++ 里
-3 % 10 = -3 ； -3 也在 -3 + 10 = 7 的同余域
3 % -10 = 3 ；3 也在 3 - 10 = -7 的同余域 -
对m个数进行查找的时候， 先查找 x % a[n] 对应的同余域的map,再使用it = lower_bound(x) ，如果找到x那么答案就是 x 对应的 i。 如果每找到，模数为正那么最合适的答案在 --it位置，否则就在 ++it位置(这个很好分析出来，可以自己分析下
code
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define IOS std::ios_base::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
const int N = 1e5 + 10, inf = 1e9 + 10;
ll a[N];
unordered_map<ll, map<ll, ll>> mp;
unordered_map<ll, ll> nmp;
int n, m, T;
int main()
{
IOS cin >> T;
while (T--)
{
mp.clear();
nmp.clear();
cin >> n >> m;
for (int i = 1; i <= n; ++i)
{
cin >> a[i];
a[i] += a[i - 1];
}
if (a[n])
{
for (int i = n; i; --i)
{
ll z = a[i] % a[n];
if (mp.count(z))
{
mp[z][a[i]] = i;
}
else
{
map<ll, ll> bmp;
bmp.insert(pair<ll, ll>(a[i], i));
mp[z] = bmp;
}
if ((a[i] > 0 && a[n] < 0) || (a[i] < 0 && a[n] > 0))
{
z += a[n];
if (mp.count(z))
{
mp[z][a[i]] = i;
}
else
{
map<ll, ll> bmp;
bmp.insert(pair<ll, ll>(a[i], i));
mp[z] = bmp;
}
}
}
while (m--)
{
ll x;
cin >> x;
if (!x)
cout << 0 << '\n';
else
{
ll y = x % a[n];
if (!mp.count(y))
{
cout << -1 << '\n';
}
else
{
map<ll, ll> &fmp = mp[y];
map<ll, ll>::iterator it = fmp.lower_bound(x);
if (it != fmp.end() && it->first == x)
{
cout << it->second << '\n';
}
else if (a[n] > 0)
{
if (it == fmp.begin())
{
cout << -1 << '\n';
}
else
{
--it;
cout << (x - it->first) / a[n] * 1LL * n + it->second << '\n';
}
}
else
{
if (it == fmp.end())
cout << -1 << "\n";
else
cout << (x - it->first) / a[n] * 1LL * n + it->second << '\n';
}
}
}
}
}
else
{
for (int i = n; i; --i)
nmp[a[i]] = i;
while (m--)
{
ll x;
cin >> x;
if (!x)
cout << 0 << '\n';
else if (nmp.count(x))
cout << (nmp.find(x))->second << '\n';
else
cout << -1 << '\n';
}
}
}
return 0;
}
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