代码搜索：Inversion
找到约 626 项符合「Inversion」的源代码
代码结果 626
www.eeworm.com/read/405817/11456379
cpp minimum inversion number(逆序数).cpp
#include
#include
#include
using namespace std;
const int MAX = 5100;
//逆序数值存放在anti中
int num[MAX];
int p[MAX], t[MAX], anti = 0;
void merge(int first, int last)
www.eeworm.com/read/300587/13904137
cpp 1394 minimum inversion number.cpp
#include
using namespace std;
int num[5001];
int imin,n;
int main()
{
int i,j,t;
while( scanf("%d",&n)!=EOF )//输入的数字范围 0 ~ n-1
{
for(i=0;i
www.eeworm.com/read/279401/10440847
readme
This is a README to the FATOMO code (Ver. Oct., 2000)
to give a brief overview how to use fatomo
describing the different input & output files.
Dr. St. Husen
Inst. o
www.eeworm.com/read/237289/13969955
readme
This is a README to the FATOMO code (Ver. Oct., 2000)
to give a brief overview how to use fatomo
describing the different input & output files.
Dr. St. Husen
Inst. o
www.eeworm.com/read/405817/11456315
cpp inversion(构造逆序数用线段树).cpp
#include
#include
#include
#include
using namespace std;
struct node {
long long l,r;
node * pl, * pr;
long long count;
}mem[150000];
long long m
www.eeworm.com/read/220554/14796700
txt usage.txt
Short guide for using the tomo3d package (S閎astien Judenherc, Mar 2002).
1/ install the package + dependencies
install libsismoutil *without* iasp support (see cvs on nilgiri.u-strasbg.fr)
install ia
www.eeworm.com/read/248950/12531294
c invspd.c
/* invspd.c: Do an "in place" inversion of a real square symmetric positive
* definite matrix "a" of size "n" by "n" and return log of it's determinant.
*
* The function only looks at elements on o
www.eeworm.com/read/109462/15557204
txt gaafilehelp.txt
This file describes the input file of the GAA program.
The Definition file is required, the others are optional
1. Definition File
The main input file is the Definition File, which is require
www.eeworm.com/read/405817/11456295
cpp minimum inversion number(线段树求逆序数).cpp
#include
#include
#include
using namespace std;
struct node {
int l,r;
node * pl, * pr;
int count;
}mem[13000];
int mem_pos;
int n,num[5100];
int anti, an