6661 Equal Sum Sets (DP), 6661 sets

6661 Equal Sum Sets (DP), 6661 sets
Let us consider sets of positive integers less than or equal to n. Note that all elements of a set are
Different. Also note that the order of elements doesnt matter, that is, both {3, 5, 9} and {5, 9, 3} mean
The same set.
Specifying the number of set elements and their sum to be k and s, respectively, sets satisfying
Conditions are limited. When n = 9, k = 3 and s = 23, {6, 8, 9} is the only such set. There may be
More than one such set, in general, however. When n = 9, k = 3 and s = 22, both {5, 8, 9} and {6, 7, 9}
Are possible.
You have to write a program that calculates the number of the sets that satisfy the given conditions.
Input
The input consists of multiple datasets. The number of datasets does not exceed 100.
Each of the datasets has three integers n, k and s in one line, separated by a space. You may assume
1 ≤ n ≤ 20, 1 ≤ k ≤ 10 and 1 ≤ s ≤ 155.
The end of the input is indicated by a line containing three zeros.
Output
The output for each dataset shocould be a line containing a single integer that gives the number of
Sets that satisfy the conditions. No other characters shoshould appear in the output.
You can assume that the number of sets does not exceed 231 −1.
Sample Input
9 3 23
9 3 22
10 3 28
16 10 107
20 8 102
20 10 105
20 10 155
3 4 3
4 2 11
0 0 0
Sample Output
1
2
0
20
1542
5448
1
0

0

DP recurrence: dp [I] [k] [s] indicates the number recurrence relationship: dp [I] [k] [s] = dp [I-1] [k] [s] + dp [I-1] [k-1] [s-I].

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000+100;int dp[21][11][156];void init(){    memset(dp,0,sizeof(dp));    dp[1][1][1]=1;    for(int i=1;i<=20;i++)    {        dp[i][1][i]=1;        dp[i][0][0]=1;    }    for(int i=2;i<=20;i++)    {        for(int k=1;k<=10;k++)        {            if(k>i)   continue;            for(int s=1;s<=155;s++)            {                int sum=0;//                for(int j=1;j<i&&j<=k;j++)                    sum+=dp[i-1][k][s];                if(s>=i) sum+=dp[i-1][k-1][s-i];                dp[i][k][s]=sum;//                if(i<=4&&s<10)//                   cout<<i<<" "<<k<<" "<<s<<" "<<sum<<endl;            }        }    }//    cout<<dp[4][2][5]<<endl;}int main(){    init();    int n,k,s;    while(~scanf("%d%d%d",&n,&k,&s)&&(n+k+s))        printf("%d\n",dp[n][k][s]);    return 0;}

Similar HDU 2861

Problem DescriptionPatti and Terri run a bar in which there are 15 stools. One day, Darrell entered the bar and found that the situation how MERs chose the stools were as follows:
OOEOOOOEEEOOOEO
O means that the stool in a certain position is used, while E means that the stool in a certain position is empty (here what we care is not who sits on the stool, but whether the stool is empty ). as the example we show above, we can say the situation how the 15 stools is used determines 7 intervals (as following ):
OO E OOOO EEE OOO E O

Now we postulate that there are N stools and M MERs, which make up K intervals. How many arrangements do you think will satisfy the condition?
 
InputThere are multi test cases and for each test case:
Each case contains three integers N (0 <N <= 200), M (M <= N), K (K <= 20 ).
OutputFor each test case print the number of arrangements as described abve. (All answers is fit in 64-bit .)
Sample Input

3 1 34 2 4

 
Sample Output

12

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>typedef long long LL;using namespace std;LL dp[201][201][22][2];void init(){    memset(dp,0,sizeof(dp));    dp[1][0][1][0]=1;    dp[1][1][1][1]=1;    for(int i=1;i<=200;i++)    {        dp[i][0][1][0]=1;        dp[i][i][1][1]=1;    }    for(int i=2;i<=200;i++)    {        for(int j=1;j<=i;j++)        {            for(int k=1;k<=20&&k<=i;k++)            {                dp[i][j][k][0]=dp[i-1][j][k-1][1]+dp[i-1][j][k][0];                dp[i][j][k][1]=dp[i-1][j-1][k][1]+dp[i-1][j-1][k-1][0];            }        }    }//    cout<<dp[12][13][15][0]<<endl;}int main(){    int n,m,k;    init();    while(~scanf("%d%d%d",&n,&m,&k))        printf("%I64d\n",dp[n][m][k][0]+dp[n][m][k][1]);    return 0;}

X ^ 4/(x ^ 3 + 1) indefinite points

Take a look at the following process. Will rational fraction be disassembled?




What are commonly used English in computers )?

Computer English Vocabulary
Access arm head arm and inventory arm
Access time
Adder
Address
Alphanumeric
Analog computer simulation computer
Analyst
Area
Array, array
Assembler
Automation
Band Area
Batch processing
Binary code
Binary digit binary bit and binary number
Bit/binary
Branch and branch line
Brush
Buffer storage buffer
Calculator
Call instruction call command
Card punch card punching machine
Card reader and reader
Cell
Channel, channel
Character
Check digit
Circuit, line
To clear and clear
Clock
Code
To code
Coder coders, encoders
Command and command
Compiler Compilation Program
Computer language
Console
Control unit control components and controllers
Core storage and core store core memory
Counter
Cybernetics Control Theory
Cycle
Data
Data processing
Debugging
Describe
Digit number, digit, bit
Digital computer
Disc, disk
Display unit display Device
Drum
To edit
Electronics
Emitter Transmitter
To encode Encoding
To erase erasure, cleaning, and erasing
Feed and Supply
To feed and Supply
Feedback
Field, Information Group, Domain
File
Floppy disk
Floppy disk drive
Flow chart Flowchart
Frame
Hardware
Identifier
Index
Information
Inline processing
Input
Inquiry
Instruction command
Integrated circuit
To interpret explanation
Item project, item
Jump Transfer
Key and key code
Keyboard
Latency time wait time
Library, library
Linkage connection
To load loading, storage, writing, and loading
Location storage unit
Logger registers and recorders
Loop
Machine language
Magnetic storage magnetic memory
Magnetic tape
Matrix
Memory
Message and message
Microcomputer
Module components and modules
Monitor, Monitoring Program,... remaining full text>