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). (
) (
)
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 q lines - the answers to the queries.
3 4
1 2 3
3
s 1 3 2
+ 2 1
- 1 2
2
3
1
题目大意：
s 求 l~r中 对m取模==mod 的 和
+ 单点修改
- 单点修改,如果减后小于0直接输出
树状数组
#include <ctype.h>
#include <cstdio>
#define N 10005
typedef long long LL;
void read(LL &x)
{
x=0;bool f=0;
char ch=getchar();
for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
x=f?(~x)+1:x;
}
LL dis[N],n,m,q;
struct node
{
LL tag[N];
int n;
int lowbit(int x) {return x&((~x)+1);}
void plus(int x,int y)
{
for(;x<=n;x+=lowbit(x)) tag[x]+=y;
}
LL query(int x)
{
LL ans=0;
for(;x;x-=lowbit(x)) ans+=tag[x];
return ans;
}
}a[25];
int main()
{
read(n);read(m);
for(int i=0;i<m;i++) a[i].n=n;
for(int i=1;i<=n;i++)
{
read(dis[i]);
a[dis[i]%m].plus(i,dis[i]);
}
char str[5];
read(q);
for(LL x,y,z;q--;)
{
scanf("%s",str+1);read(x);read(y);
switch(str[1])
{
case 's':
{
read(z);
printf("%lld\n",a[z].query(y)-a[z].query(x-1));
break;
}
case '+':
{
a[dis[x]%m].plus(x,-dis[x]);
dis[x]+=y;
a[dis[x]%m].plus(x,dis[x]);
printf("%lld\n",dis[x]);
break;
}
case '-':
{
if(dis[x]<y) {printf("%lld\n",dis[x]);}
else
{
a[dis[x]%m].plus(x,-dis[x]);
dis[x]-=y;
a[dis[x]%m].plus(x,dis[x]);
printf("%lld\n",dis[x]);
}
break;
}
}
}
return 0;
}
