codeforces 630~~【组合数 C(n,m)】

F - Selection of Personnel

Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received n resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate.

Input

The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes.

Output

Output one integer — the number of different variants of group composition.

Sample Input

Input

7

Output

29

组合数 ： 求 C(n,5)+C(n,6)+C(n,7)

#include <cstdio>  #include <cstring>  #include <cmath>  #include <cstdlib>  #include <algorithm>  #include <queue>  #include <stack> #define INF 0x3f3f3f3f  #define eps 1e-8   #define Si(a) scanf("%d", &a)  #define Sl(a) scanf("%lld", &a)  #define Sf(a) scanf("%lf", &a)  #define Ss(a) scanf("%s", a)  #define Pi(a) printf("%d\n", (a))  #define Pf(a) printf("%.2lf\n", (a))  #define Pl(a) printf("%lld\n", (a))  #define Ps(a) printf("%s\n", (a))  #define Wi(a) while((a)--)  #define cle(a, b) memset(a, (b), sizeof(a))  #define MOD 1000000007  #define LL long long  #define PI acos(-1.0) using namespace std;__int64 comb(__int64 n, int m){__int64 a, b;a = b = 1;while(m--){a *= (n-m);a /= b;++b;}return a;}int main(){__int64 n;while(scanf("%I64d", &n)==1){__int64 ans = 0;for(int i = 5; i <= 7; ++i){ans += comb(n, i);}printf("%I64d\n", ans);}return 0;}

G - Challenge Pennants

Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Because of budget cuts one IT company established new non-financial reward system instead of bonuses.

Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.

Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.

In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.

One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.

Output

Output one integer — the amount of ways to place the pennants on n tables.

Sample Input

Input

2

Output

24

题解  

5面A锦旗发放方案为  11111，1112，122，113，14，23,5 
方案数为  sum1=C(n,5)+4*C(n,4)+2*3*C(n,3)+2*2*C(n,2)+C(n,1) 
3面B锦旗发放方案为  111，12，3 
方案数为  sum2=C(n,3)+2*C(n,2)+C(n,1) 
总方案数为ans=sum1*sum2 

#include <cstdio>  #include <cstring>  #include <cmath>  #include <cstdlib>  #include <algorithm>  #include <queue>  #include <stack> #define INF 0x3f3f3f3f  #define eps 1e-8   #define Si(a) scanf("%d", &a) #define Si64(a) scanf("%I64d", &a)  #define Sl(a) scanf("%lld", &a)  #define Sf(a) scanf("%lf", &a)  #define Ss(a) scanf("%s", a)  #define Pi(a) printf("%d\n", (a))  #define Pi64(a) printf("%I64d\n", (a)) #define Pf(a) printf("%.2lf\n", (a))  #define Pl(a) printf("%lld\n", (a))  #define Ps(a) printf("%s\n", (a))  #define Wi(a) while((a)--)  #define cle(a, b) memset(a, (b), sizeof(a))  #define MOD 1000000007  #define LL long long  #define PI acos(-1.0) #define prt(x, y)  printf("%.12lf %.12lf\n", x*fy+y*fx+px, -x*fx+y*fy+py);using namespace std;long long comb(int n, int m){LL a, b;a = 1; b = 1;while(m--){a *= (n-m);a /= b;b++;}return a;}int main(){int n;while(Si(n)==1){LL sum1 = comb(n,5)+4*comb(n,4)+2*3*comb(n,3)+2*2*comb(n,2)+comb(n,1);LL sum2 = comb(n,3)+2*comb(n,2)+comb(n,1);Pl(sum1*sum2);}return 0;}

H - Benches

Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

The city park of IT City contains n east to west paths and n north to south paths. Each east to west path crosses each north to south path, so there are n² intersections.

The city funded purchase of five benches. To make it seems that there are many benches it was decided to place them on as many paths as possible. Obviously this requirement is satisfied by the following scheme: each bench is placed on a cross of paths and each path contains not more than one bench.

Help the park administration count the number of ways to place the benches.

Input

The only line of the input contains one integer n (5 ≤ n ≤ 100) — the number of east to west paths and north to south paths.

Output

Output one integer — the number of ways to place the benches.

Sample Input

Input

5

Output

120

n*n个路口放 5 个凳子，并且每个路口只能放一个。求共有几种方式

#include <cstdio>  #include <cstring>  #include <cmath>  #include <cstdlib>  #include <algorithm>  #include <queue>  #include <stack> #define INF 0x3f3f3f3f  #define eps 1e-8   #define Si(a) scanf("%d", &a) #define Si64(a) scanf("%I64d", &a)  #define Sl(a) scanf("%lld", &a)  #define Sf(a) scanf("%lf", &a)  #define Ss(a) scanf("%s", a)  #define Pi(a) printf("%d\n", (a))  #define Pi64(a) printf("%I64d\n", (a)) #define Pf(a) printf("%.2lf\n", (a))  #define Pl(a) printf("%lld\n", (a))  #define Ps(a) printf("%s\n", (a))  #define Wi(a) while((a)--)  #define cle(a, b) memset(a, (b), sizeof(a))  #define MOD 1000000007  #define LL long long  #define PI acos(-1.0) #define prt(x, y)  printf("%.12lf %.12lf\n", x*fy+y*fx+px, -x*fx+y*fy+py);using namespace std;long long comb(int n, int m){LL a, b;a = 1; b = 1;while(m--){a *= (n-m);a /= b;b++;}return a;}int main(){int n;while(Si(n)==1){Pl(120*comb(n,5)*comb(n,5));}return 0;}