best book for permutation and combination

Question: n, m, there are n kinds of items. Each item has a [I], and the number of M items must be selected. Exponential functions, G = A0 + A1/1! * X + A2/2! * X ^ 2 +... + AK/K! * X ^ K...Details about exponential functions #include Hdu1521 permutation and combination (exponential functions)

for red. Output descriptionFor each set of test data, the output answer1000000007 ({10}^{9}+7)0000 00007 (1 0 ? 9 ?? +7) modulo. Input sample31 0 041 0 0 1Output sample48Problem : interval dp+ Combinatorial counting problems, the transfer equation is, each time you select the last position of the current interval, and then multiply the combined number c[interval length [left interval length]Note the initialization of the DP array-1//Meek///#include #include #include#include#include#include

need to solve. OutputFor each set of test data, the output line contains an integer x that represents the result of the number of spanning trees of the N-five-angle ring after modulo 2007.Sample Input12Sample Output40HINTSourceSolutionPursuit without $matrix$-$Tree $ of Pleasure ~First, each Pentagon must delete an edge, so that a base ring tree is left.Drawing can be found in this ring on the edge of the cut will lead to have a Pentagon more delete an edgeAnd if a Pentagon is deleted two sides

,j; intsum; intss=0; CharA; intb; scanf ("%d",s); intpp=0; while(cin>>aa!='Q') {miss[pp].x=A; scanf ("%d",b); MISS[PP++].y=b; SS=ss+b; intt=0; for(i=0; i) { if(q[i].aa==a) {q[i].bb+=b; T=1; } } if(t==0) {Q[num].aa=A; Q[num++].bb=b; } } intPer=1; for(i=1; i) per=per*i; if(s-ss-num+10) cout0Endl; Else{DFS (0,0,0); /*cout*/ inta[Ten]; for(intIi=0; ii) { for(intI=0; i) A[i]=i; intflag=0; for(intI=0; I) { /

Alas, I don't know what the first nerve is. The first time I defined t as char.The number of decimals is not much more 、、、 find half an hour to find the future to be carefulThe large number must be saved with a string, and a long long must overflow.The specific idea is to divide the factorial n! of all characters by the number of occurrences of each character.#include HDU 1261 permutation combination large

, how many of the 01 sequences that satisfy the condition.Observing the appearance of 1, we consider whether this occurrence can be placed in the second row, apparently, those 0,1 corresponding to this 1, either on the left side of this 1, or in front of this 1. And certainly have a 0, in front of this 1, Count 1 and 0 before this 1. That is, the requirement, The number of 0 is greater than the number of 1. OK, the problem has been resolved. Similar to parentheses. or 0 as the stack, 1 look at t

Permutations IIGiven A collection of numbers that might contain duplicates, return all possible unique permutations.For example,[1,1,2] has the following unique permutations:[1,1,2], [1,2,1], and [2,1,1].Idea: The problem is to remove duplicates compared to the previous question. The code is slightly different from the above question, the specific code is as follows:public class Solution {boolean[] b; ListHowever, the above code is inefficient when the data volume is large, such as the data h

Print the full array of stringsThe idea of the algorithm:Divides a string into two parts, the first character + all the characters in the back part. This allows the entire process to be solved recursively:1. Each character is given the first character 2. When a character is the first character, the whole arrangement of all the following characters is used, and the whole arrangement of all the characters behind it can be regarded as the sub-problem of recursion.Full-lined recursive tree:But here'

PublicClasssolution{Public listInt[] arr) {listNew ArraylistCornerif (arr = =null | | Arr.length = =0) {return res;}CoreListNew ArraylistHelper (res, curlist, arr);return res;}PublicvoidHelper(listInt[] arr) {Baseif (curlist.size () = = Arr.length) {Res.add (New Arraylistreturn;}Currentfor (int i =0; i if (Curlist.contains (Arr[i])) {continue;}Curlist.add (Arr[i]);//nextHelper (res, curlist, arr);Curlist.remove (Curlist.size ()- 1);}}}Permutation and

) scanf ("%d", (n))#defineSII (n,m) scanf ("%d%d", (N), (M))#defineCLE (A,val) memset (A, (Val), sizeof (a))#defineRep (i,b) for (int i=0;i#defineREP (I,A,B) for (int i= (a); i#defineREReP (I,A,B) for (int i= (a); i>= (b); i--)Const Doubleeps= 1e-9 ;/*/////////////////////////c o d i n g S p a C e/////////////////////////*/Const intmaxn=100000+5 ;Const intmod=1e9+7; ll JIECH[MAXN];//an array of factorial modulointn,m;//Quick Power Templatell Mod_pow (ll x,ll n,ll MoD) {ll res=1; while(n>0){

function Docombination (arr) {varCount = Arr.length-1;//array length (starting from 0) varTMP = []; varTotalarr = [];//Total Group returnDocombinationcallback (arr,0);//start with the first one//JS has no static data, in order to avoid confusion with external data, need to use the form of closuresfunction Docombinationcallback (arr, curr_index) { for(Val of Arr[curr_index]) {Tmp[curr_index]= Val;//add an array with Curr_index as an index//The current loop subscript is less than the total l