题目链接 https://atcoder.jp/contests/abc153/tasks/abc153_f

题意 ： 在坐标轴上有一些怪兽，每个怪兽有对应的生命值hi，你可以对他们进行炮击，你的每次炮击可以队该点前后D范围内的怪兽造成A的伤害，问最少要炮击多少次。

我的最初的想法是先排序，扫到最左边的怪兽，先进行炮击，把他打死，然后记录炮击了多少次，然后把其后2d距离的怪兽都炮击一下发现超时

代码如下：

inline ll read(){
   ll s=0,w=1;
   char ch=getchar();
   while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)w=-1;ch=getchar();}
   while(ch>=‘0‘&&ch<=‘9‘) s=s*10+ch-‘0‘,ch=getchar();
   return s*w;
}
const int N = 2e5+5;
struct mon
{
        ll x,h;
};
mon monster[N];
bool cmp(mon a, mon b)
{
    return a.x < b.x;
}
int main()
{
    //ios::sync_with_stdio(false);cin.tie(NULL);
    ll n, d, a;
    n = read();
    d = read();
    a = read();
    for (int i = 0; i < n; i++)
    {
        monster[i].x = read();
        monster[i].h= read();
    }
    sort(monster, monster + n,cmp);
    ll ans = 0;
    for (int i = 0; i < n; i++)
    {
        if (monster[i].h>0)
        {
            ll xx = monster[i].x + d;
            ll num = (monster[i].h + a - 1) / a;
            ans += num;
            monster[i].h = 0;
            bool flag = false;
            for (int j = i + 1; monster[j].x >= (xx - d) && monster[j].x<= xx + d && j < n;j++)
            {
                monster[j].h-=a*num;

            }
        }
    }
    cout << ans << endl;

}

后来学习了大佬的代码，他是这么操作的，建立一个数组c[i]，记录每个点已经炮击了几次need，则c[i] += need,下一个点已经炮击的次数就是c[i + 1] += c[i],因为范围外就无效，所以每个点i你要先找到他最远能打到哪里，然后要事先减掉need，即c[j] -= need; 这样，当i到了j-1时，执行c[i+1] += c[i], need正好抵消，我觉得这个技巧太妙了

代码：

#include<iostream>
#include<string>
#include <cstdlib>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include <iomanip>

// #pragma comment(linker, "/STACK:1024000000,1024000000")
// #define pi acos(-1)
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x7f7f7f7f //2139062143
#define INF1 0x3f3f3f3f //1061109567
#define INF2 2147483647
#define llINF 9223372036854775807
#define pi 3.141592653589793//23846264338327950254
#define pb push_back
#define ll long long
#define debug cout << "debug\n";
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// ios::sync_with_stdio(false);cin.tie(NULL);
#define scai(x) scanf("%d", &x)
#define sca2i(x, y) scanf("%d %d", &x, &y)
#define scaf(x) scanf("%lf", &x)
#define sca2f(x, y) scanf("%lf %lf", &x, &y)
#define For(m,n) for (int i = m;  i < n; i++)

#define local
#ifdef local
#endif

#define MAX 10233
#define LCH(i) ((i) << 1)
#define RCH(i) ((i) << 1 | 1)
inline ll read(){
   ll s=0,w=1;
   char ch=getchar();
   while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)w=-1;ch=getchar();}
   while(ch>=‘0‘&&ch<=‘9‘) s=s*10+ch-‘0‘,ch=getchar();
   return s*w;
}
const int N = 2e5+5;
ll c[N];
struct mon
{
        ll x,h;
};
mon monster[N];
bool cmp(mon a, mon b)
{
    return a.x < b.x;
}
int main()
{
    //ios::sync_with_stdio(false);cin.tie(NULL);
   // cout << 50ll << endl;
    ll n, d, a;
    n = read();
    d = read();
    a = read();
    for (int i = 0; i < n; i++)
    {
        monster[i].x = read();
        monster[i].h= read();
    }
    sort(monster, monster + n,cmp);
    ll ans = 0;
    for (int i = 0,j = 0; i < n; i++)
    {
        while(j < n && monster[j].x <=monster[i].x + 2*d)
            ++j;
        ll need = max((monster[i].h - c[i]*a + a - 1)/a, 0ll);
        ans += need;
        c[i] += need;
        c[j] -= need;
        c[i+1] += c[i];
    }
    cout << ans << endl;

}

原文地址：https://www.cnblogs.com/hulian425/p/12236097.html