传送门
题目
Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my friend, a mathematician, thinks that it is very fun to play with a sequence of integer numbers. He writes the sequence in a row. If he wants he increases one number of the sequence, sometimes it is more interesting to decrease it (do you know why?..) And he likes to add the numbers in the interval 【l;r】. But showing that he is really cool he adds only numbers which are equal some mod (modulo m).
Guess what he asked me, when he knew that I am a programmer? Yep, indeed, he asked me to write a program which could process these queries (n is the length of the sequence):
+ p r It increases the number with index p by r. (, )
You have to output the number after the increase.
- p r It decreases the number with index p by r. (, ) You must not decrease the number if it would become negative.
You have to output the number after the decrease.
s l r mod You have to output the sum of numbers in the interval which are equal mod (modulo m). () ()
Input
The first line of each test case contains the number of elements of the sequence n and the number m. (1?≤?n?≤?10000) (1?≤?m?≤?10)
The second line contains n initial numbers of the sequence. (0?≤?number?≤?1000000000)
The third line of each test case contains the number of queries q (1?≤?q?≤?10000).
The following q lines contains the queries (one query per line).
Output
Output q lines - the answers to the queries.
题目大意
给你n和m以及n个数，q次操作，操作有三种：
在第p个数加r，输出现在的值
如果第p个数减r非负则将其减r，输出这个数
查询lr区间内模m余mod的数的和
分析
运用分块，因为m很小，我们可以记录每块中模m的各个可能值各种之和，同时维护每个数的值，修改是要将原来所在的分组减去原来数，再在新分组加上这个数。
代码
#include
#include
#include
#include[span style="color: rgba(0, 0, 255, 1)">string
#include
#include
#include
#include
#include
#include
#include
#include[span style="color: rgba(0, 0, 255, 1)">set
#include
#include
using namespace std;
long long n,m,sum,block,all【11000】【20】,beg【11000】,a【11000】,q,r【11000】,l【11000】;
inline long long read(){
long long x=0,f=1;char s=getchar();
while(s[span style="color: rgba(128, 0, 0, 1)">'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0's<='9'){x=(x[3)+(x[1)+(s-'0');s=getchar();}
return x*f;
}//代码效果参考：http://www.ezhiqi.com/bx/art_3125.html
inline void init(){
long long i,j,k;
block=sqrt(n);
sum=n%block==0?n/block:n/block+1;
for(i=1;i<=sum;i++)
l【i】=r【i-1】+1,r【i】=r【i-1】+block;
r【sum】=n;
for(i=1;i<=n;i++){
beg【i】=(i-1)/block+1;
all【beg【i】】【a【i】%m】+=a【i】;
}
}
inline void go(long long x,long long y){
if(a【x】+y[span style="color: rgba(128, 0, 128, 1)">0){
printf("%I64d\n",a【x】);
return;
}
all【beg【x】】【a【x】%m】-=a【x】;
a【x】+=y;
all【beg【x】】【a【x】%m】+=a【x】;
printf("%I64d\n",a【x】);
}
inline void work(long long x,long long y,long long mod){
long long i,j,k,ans=0;
if(beg【x】==beg【y】){
for(i=x;i<=y;i++)
if(a【i】%m==mod)ans+=a【i】;
}else {
for(i=x;i<=r【beg【x】】;i++)
if(a【i】%m==mod)ans+=a【i】;
for(i=beg【x】+1;i
ans+=all【i】【mod】;
for(i=l【beg【y】】;i<=y;i++)
if(a【i】%m==mod)ans+=a【i】;
}
printf("%I64d\n",ans);
return;
}
int main(){
long long i,j,k,x,y,mod;
n=read(),m=read();
for(i=1;i<=n;i++)a【i】=read();
init();
q=read();
for(i=1;i<=q;i++){
char c;
cinc;
if(c=='+'){
x=read(),y=read();
go(x,y);
}else if(c=='-'){
x=read(),y=read();
go(x,-y);
}else {
x=read(),y=read(),mod=read();
work(x,y,mod);
}
}
return 0;
}//代码效果参考：http://www.ezhiqi.com/zx/art_2404.html
