CODECHEF Sereja and Commands-优快云博客

本文链接：https://blog.youkuaiyun.com/wry0112/article/details/77936981

本文提供了一道CodeChef竞赛题目SEACO的解决方案，利用树状数组进行区间更新和查询操作，通过倒序处理和取模注意事项实现了有效的解题策略。

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题目链接：https://www.codechef.com/SEPT17/problems/SEACO
树状数组，倒着处理
因为数字更新可能为负，所以取模要注意
贴代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=100005,P=1000000007;
int b[maxn],c[maxn];
int n,m,Q;
struct rc{
    int t,l,r;
}a[maxn];
int lowbit(int x){
    return x&(-x);
}
void data_1(int x,int y){
    for (int i=x;i<=m;i+=lowbit(i))b[i]=(ll(b[i])+ll(y)+ll(P))%ll(P);
}
void data_2(int x,int y){
    for (int i=x;i<=n;i+=lowbit(i))c[i]=(ll(c[i])+ll(y)+ll(P))%ll(P);
}
int query_1(int x){
    int s=0;
    for (int i=x;i>=1;i-=lowbit(i))s=(ll(s)+ll(b[i])+ll(P))%ll(P);
    return s;
}
int query_2(int x){
    int s=0;
    for (int i=x;i>=1;i-=lowbit(i))s=(ll(s)+ll(c[i])+ll(P))%ll(P);
    return s;
}
int main(){
//  freopen("5.in","r",stdin);
//  freopen("5.out","w",stdout);
    scanf("%d",&Q);
    while (Q--){
        scanf("%d%d",&n,&m);
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        for (int i=1;i<=m;i++)scanf("%d%d%d",&a[i].t,&a[i].l,&a[i].r);
        for (int i=m;i>=1;i--){
            if (a[i].t==1){
                int x=query_1(i)+1;
                data_2(a[i].l,x);data_2(a[i].r+1,-x);
            }else{
                int x=query_1(i)+1;
                data_1(a[i].l,x);data_1(a[i].r+1,-x);
            }
        }
        for (int i=1;i<=n;i++)printf("%d ",query_2(i));
        printf("\n");
    }
    return 0;
}