【Loj 6280】区间加区间求和

最新推荐文章于 2020-11-23 20:08:11 发布

原创最新推荐文章于 2020-11-23 20:08:11 发布 · 198 阅读

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本文详细解析了Loj6280题目的解题思路，通过使用sum数组来维护区间的和，实现单点修改与区间查询功能。文章分享了作者的解题过程，包括如何避免WA，以及给出完整的C++代码实现。

Loj 6280
题意其实这题就是第一题拿来改一下
很骚的一点就是我们用一个sum数组去维护每个段的和
单点修改的时候sum数组也需要改
否则sum数组不需要改一开始没注意到这个地方就WA了

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 50025;
int pos[MAX_N],col[MAX_N],arr[MAX_N],block,sum[MAX_N];
void add(int x,int y,int v)
{
    int st = pos[x] + 1,ed = pos[y] - 1;
    for(int i = x;i<=min(pos[x]*block,y);++i)
        arr[i] += v,sum[pos[x]]+=v;
    if(pos[x]!=pos[y])
        for(int i = (pos[y]-1)*block+1;i<=y;++i)
            arr[i]+= v,sum[pos[y]]+=v;
    for(int i = st;i<=ed;++i)
        col[i] += v;
}
ll query(int x,int y,int v)
{
    ll ans = 0;
    int st = pos[x] + 1,ed = pos[y] - 1;
    for(int i = x;i<=min(pos[x]*block,y);++i)
        (ans+=arr[i]+col[pos[x]])%=(v+1);
    if(pos[x]!=pos[y])
        for(int i = (pos[y]-1)*block+1;i<=y;++i)
            (ans+=arr[i]+col[pos[y]])%=(v+1);
    for(int i = st;i<=ed;++i)
        (ans+=(1ll*block*col[i]+sum[i]))%=(v+1);
    return ans;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,opt,x,y,v;scanf("%d",&n);
    block = sqrt(n);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),pos[i] = (i-1)/block+1,sum[pos[i]]+=arr[i];
    for(int i = 1;i<=n;++i)
    {
        scanf("%d%d%d%d",&opt,&x,&y,&v);
        if(opt==0) add(x,y,v);
        else printf("%lld\n",query(x,y,v));
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}