2021CCPC重赛 1005 Monopoly
作者:互联网
2021CCPC重赛 1005 Monopoly
Solution
注意对负数的取模方法,注意特判0
查找最后一个小于等于x的方法:
使用 pair <元素值乘 -1 ,数组内下标>,排序后查找第一个大于等于 -x 的值
代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<math.h>
#include<queue>
#include<set>
//#define int long long
using namespace std;
typedef pair<int, int> pii;
typedef double dd;
typedef long long ll;
const int MAXN = 100010;
const int MAXM = 400010;
const dd eps = 1e-6;
int n, m;
ll a[MAXN], sum[MAXN];
map<ll, vector<pair<ll, int>>> mp;
void solve(int flag)
{
for (int i = 1; i <= n;i++)
{
ll id = (sum[i] % sum[n] + sum[n]) % sum[n];
mp[id].push_back({-sum[i], i});
}
for (auto now = mp.begin(); now != mp.end();now++)
{
sort(now->second.begin(), now->second.end());
}
while (m--)
{
ll x;
scanf("%lld", &x);
if(x == 0)
{
printf("0\n");
continue;
}
x *= flag;
ll xx = (x % sum[n] + sum[n]) % sum[n];
if (mp.count(xx) == 0)
{
printf("-1\n");
continue;
}
pair<ll, int> p = {-x, 0};
int id = lower_bound(mp[xx].begin(), mp[xx].end(), p) - mp[xx].begin();
if (id == (int)mp[xx].size())
printf("-1\n");
else
{
ll ans = (x + mp[xx][id].first) / sum[n] * n + (ll)mp[xx][id].second;
printf("%lld\n", ans);
}
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
memset(sum, 0, sizeof(sum));
mp.clear();
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]), sum[i] = sum[i - 1] + a[i];
if(sum[n] > 0)
{
solve(1);
}
else if(sum[n] < 0)
{
for (int i = 1; i <= n;i++)
sum[i] = -sum[i];
solve(-1);
}
else if(sum[n] == 0)
{
map<ll, int> mpp;
for (int i = 1; i <= n;i++)
{
ll id = sum[i];
if(mpp.count(id) == 0)
mpp[id] = i;
}
while(m--)
{
ll x;
scanf("%lld", &x);
if(x == 0)
{
printf("0\n");
continue;
}
if(mpp.count(x) == 1)
printf("%d\n", mpp[x]);
else
printf("-1\n");
}
}
}
}
标签:ll,int,sum,2021CCPC,mp,1005,include,id,重赛 来源: https://blog.csdn.net/weixin_53177273/article/details/120811822