[洛谷P3962]浮游大陆的68号岛

最新推荐文章于 2024-07-12 22:30:57 发布

原创最新推荐文章于 2024-07-12 22:30:57 发布 · 396 阅读

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本文介绍了一种利用两次前缀和算法解决特定问题的方法，通过距离、总价值及个数的前后缀维护，实现复杂查询的有效处理。文章提供了详细的算法实现代码，包括取模操作下的负数处理技巧。

题目←

两次前缀和来解决，式子并不难推
将二次前缀想像成三角形会比较好理解

需维护：
1、距离前缀；
2、总价值(运费)前后缀；
3、个数前后缀；

具体式子详见代码
比较恶心的是取模意义下随时会有负数，并不太会处理，求大佬指教：
qz、hz中的1表示运费，2表示个数

#include<iostream>
#include<cstdio>
#include<algorithm>
#define LL long long
#define P 19260817
using namespace std;
const int MAXN = 200000 + 50;
LL qz1[MAXN],qz2[MAXN],hz1[MAXN],hz2[MAXN];
LL num[MAXN],dis[MAXN],ju[MAXN];
LL n,x,tot,m;
int l,r,v;
int main(){
    scanf("%lld%lld",&n,&m);
    for(int i = 2;i <= n;i ++){
        scanf("%lld",&ju[i]);
        dis[i] = (dis[i - 1]%P + ju[i]%P)%P;
    }
    for(int i = 1;i <= n;i ++){
        scanf("%lld",&num[i]);
        qz1[i] = (qz1[i - 1]%P + (tot%P*ju[i]%P)%P)%P;
        tot = (num[i]%P + tot%P)%P;
        qz2[i] = tot;
    }
    tot = 0;
    for(int i = n;i >= 1;i --){
        hz1[i] = (hz1[i + 1]%P + (tot%P*ju[i + 1]%P)%P)%P;
        tot = (num[i]%P + tot%P)%P;
        hz2[i] = tot;
    }
    /*for(int i = 1;i <= n;i ++){
        printf("%lld ",hz1[i]);
    }*/
    for(int i = 1;i <= m;i ++){
        scanf("%d%d%d",&v,&l,&r);
        LL ans = 0;
        if(v >= l && v <= r){
            LL temp1 = qz1[v] - qz1[l];
            while(temp1 < 0)temp1 += P;
            LL tmp = dis[v] - dis[l];
            while(tmp < 0)tmp += P;
            temp1 = (temp1 - qz2[l - 1]%P*(tmp%P))%P;
            while(temp1 < 0)temp1 += P;
            while(tmp < 0)tmp += P;
            LL temp2 = hz1[v] - hz1[r];
            while(temp2 < 0)temp2 += P;
            tmp = dis[r] - dis[v];
            while(tmp < 0)tmp += P;
            temp2 = (temp2 - hz2[r + 1]%P*(tmp%P))%P;
            while(temp2 < 0)temp2 += P;
            ans = (temp1%P + temp2%P)%P;
        }
        else if(v > r){
            LL tmp = dis[r] - dis[l];
            while(tmp < 0)tmp += P;
            LL tmp2 = qz1[r] - qz1[l];
            while(tmp2 < 0)tmp2 += P;
            LL temp1 = tmp2 - (qz2[l - 1]%P*tmp)%P;
            while(temp1 < 0)temp1 += P;
            tmp = dis[v] - dis[r];
            while(tmp < 0)tmp += P;
            tmp2 = qz2[r] - qz2[l - 1];
            while(tmp2 < 0)tmp2 += P;
            temp1 = (temp1 + (tmp2*tmp)%P)%P;
            ans = temp1;
        }
        else if(v < l){
            LL tmp = dis[r] - dis[l];
            while(tmp < 0)tmp += P;
            LL tmp2 = hz1[l] - hz1[r];
            while(tmp2 < 0)tmp2 += P;
            LL temp1 = tmp2 - (hz2[r + 1]*(dis[r] - dis[l])%P)%P;
            while(temp1 < 0)temp1 += P;
            tmp = dis[l] - dis[v];
            while(tmp < 0)tmp += P;
            tmp2 = hz2[l] - hz2[r + 1];
            while(tmp2 < 0)tmp2 += P;
            temp1 = (temp1 + (tmp2*tmp)%P)%P;
            ans = temp1;
        }
        printf("%lld\n",ans);
    }
    return 0; 
}