Nanyang OJ Soldier kills (ii) title NYOJ data structure



/* Soldier kills (ii)
Time limit: Ms | Memory Limit: 65535 KB
Difficulty: 5
Describe
South Generals have n soldiers, numbered 1 to N, each of these soldiers is known for their kills.

The staff is the South General's military advisers, the South General often want to know the number m to nth soldier's total kills, please help the handyman to answer the South General.

The South General's inquiry after the soldier I may also kill the Q person, after the South General asked again, need to consider the new kill number.

Input
Only one set of test data
The first line is two integer n,m, where N represents the number of soldiers (1<n<1000000), and M represents the number of instructions. (1<m<100000)
The next line is n integers, and AI represents the number of soldier kills. (0<=ai<=100)
The subsequent M-line is an instruction that contains a string and two integers, first a string, and if the string query indicates that the South General has a query operation, followed by the two integer m,n, indicating the start and end of the query soldier number; If the string add is followed by the two integer I , A (1<=i<=n,1<=a<=100), which indicates that the number of new kills for the first soldier is a.

Output
For each query, output an integer r that represents the total number of kills of soldier number m to nth soldier, one row per set of outputs
Sample input
5 6 1 2 3 4 5 Query 1 3 Add 1 2 Query 1 3 Add 2 3 Query 1 2 query 1 5
Sample output
6 8 8 20
Code: */
Normal timeout code
#include <stdio.h>
#include <stdlib.h>
#define N 1000001
int s[n],m,n,k,l;
void Ji ()
{
for (int i=k;i<=n;i++)
S[i]+=l;
}
void Chu ()
{printf ("%d\n", S[l]-s[k-1]);}
int main ()
{
scanf ("%d%d", &n,&m);
s[0]=0;
for (int i=1;i<=n;i++)
{
scanf ("%d", &s[i]);
S[I]+=S[I-1];
}
Char h[100];
for (int j=0;j<m;j++)
{
scanf ("%s%d%d", h,&k,&l);
if (h[0]== ' Q ') printf ("%d\n", s[l]-s[k-1]);
if (h[0]== ' A ') Ji ();
}
return 0;
}
Use a tree array to solve the problem of changing arrays dynamically, because a soldier in front changes the number of kills, followed by changes
#include <stdio.h>
int a[1000010],n;
int lowbit (int pat)
{
Return pat& (-PAT);//The return value is a K-power of 2, and K is the number of the Pat converted to 2 binary from right to left 0
}
void pus (int pat,int numb)
{
while (pat<=n)
{
A[pat]+=numb;
Pat+=lowbit (PAT);
}
}
int getsum (int pat)
{
int sum=0;
while (pat>=1)
{
SUM+=A[PAT];
Pat-=lowbit (PAT);
}
return sum;
}
int main ()
{
int m,i,from,to,num;
Char str[10];
scanf ("%d%d", &n,&m);
for (I=1;i<=n;++i)
{
scanf ("%d", &num);
Pus (i,num);
}
while (m--)
{
scanf ("%s%d%d", str,&from,&to);
if (str[0]== ' Q ')
printf ("%d\n", Getsum (To)-getsum (from-1));
Else
{
Pus (from,to);
}
}
return 0;
}

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Nanyang OJ Soldier kills (ii) title NYOJ data structure