msrp 1060

1060. is they Equal (25)If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 is considered equal since they is B Oth saved as 0.123*105 with simple chopping. Now given the number of significant digits in a machine and both float numbers, you is supposed to tell if they is treate D equal in this machine.Input Specification:Each input file contains one test case which gives three numbers N, A and B, where N (Output Specif

Problem description ============ SQL Server 2000 installation failed Troubleshooting ============ The following error message exists in sqlstp. log: ================ 17:02:11 end action: updatesystempath 17:02:12 C:/program files/Microsoft SQL Server/80/tools/binn/cnfgsvr.exe-F "C:/Windows/sqlstp. log "-I MSSQLServer-V 1-M 0-Q" SQL _latin1_general_cp1_ci_as "-H 721788-U sa-P 17:02:12 process exit codes: (193) 17:02:23 setup failed to configure the server. Refer to the server error logs and C:/Wi

Greedy.#include #include#include#include#defineMAXV 500500#defineMaxe 1000500using namespacestd;intN,s,x,y,z,f[maxv],g[maxv],nume=1;Long Longans=0;structedge{intV,W,NXT;} E[maxe];voidAddedge (intUintVintW) {e[++nume].v=v; E[NUME].W=W; E[NUME].NXT=G[u]; G[u]=Nume;}voidDfsintXintFath) { for(intI=g[x];i;i=e[i].nxt) { intv=e[i].v; if(V==fath)Continue; DFS (V,X); F[x]=max (f[x],f[v]+E[I].W); } for(intI=g[x];i;i=e[i].nxt) { intv=e[i].v; if(V==fath)Continue; Ans+=f[x]-f[v]-E[I].W;

^"Ans[a].idx;}intMain () {scanf ("%d",N); CIN>>s>>T; S=Qu (s); T=qu (t); if(s[0]=='0's[1]=='.') { intnum=0; for(intI=2; I) { if(s[i]=='0') num++; Else{ans[0].idx=-num; ans[0].f="0."; for(intJ=i; J0].f=ans[0].f+S[j]; Break; } } if(ans[0].f.length () = =0) {ans[0].f="0.0"; ans[0].idx=0; } } Else { intpos=-1; for(intI=0; Iif(s[i]=='.') pos=i; if(pos!=-1) {ans[0].f="0."; for(intI=0; Iif(s[i]!='.') ans[0].f=ans[0].f+R[i]; ans[0].idx=POS;

#include#include#include#includeUsingnamespace Std;int n;String Deal (String S,int e)"Thinking" 1: Absorbing excess 0:. 2: Find "." Judged with a size of 1. 3: Remove "." The 10 index (plus or minus is related to Step2) {4: To determine whether the deletion is finished, the deletion indicates that the number is 0. 5: Deposit The first n numbers, less than 0 complement. Returnint k=0;while (S.size () >0 s[0]==' 0 ')Absorption. 0 S.erase (S.begin ()) in front of the excess;if (s[0]==‘.‘)Situation

Leftmost DigitTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Problem Descriptiongiven A positive integer N, you should output the leftmost digit of n^n.Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.Each test case is contains a single positive integer N (1Outputfor Each test case, you should output the leftmost digit of n^n.Sample Input234 Sample Output22Analysis:

Topic: Given a tree has a root, each operation can make an edge right +1, the minimum number of operations, so that the root node to each leaf node distance is equalTree-shaped DPAn easy-to-find operation is more advantageous for edges closer to the root nodeFirst, sweep through each node the maximum distance from the node to all the leaf nodes in the subtree and enumerate each son to make the Benquan between the node and the son equal to the maximum distance.This is done for each node to ensure

the highest cardinality.To x^x the logarithm, x* ln (x)/ln (10), it is assumed that the value is X.ABCDEEFG so 10^x is the highest corresponding weight, 10^ 0.ABCDEFG is the highest cardinality. Note that what you get here is not an integer, why? Because here is the force of the next bit of the value is also converted to the highest level up, which is a bit like the large number, if the system is not satisfied with the forced carry, it is obvious that it will be in the high one decimal rather t

Set A to a prime number, modulus to another prime, and then the answer to the polynomial of violence, and if the answer is equal, the two polynomial are considered equal.This kind of hash has the error probability question why still must use the hash? Because the probability of error is too small, a and modulus of the value of the person can not get a good card.And then contributed 2 times WA, for the first time because the number is not judged by the boundary, the second time because the multip

#include using namespace STD; vector vectorchar> >Vintn,m,use[ -],cnt[ -];Charans[ -],s[Ten];intTopsort (intx) {inti,j,t[ -],f=1, R,c;memset(ANS,0,sizeof(ans)); for(i=0; i -; i++) T[i]=cnt[i]; C=0; R=0; while(x--) {c=0; for(i=0; iif(t[i]==0) {j=i; C + +; } }if(c>=1) {if(c>1) f=0; for(i=0; I' A '; t[j]=-1; ans[r]=0; }Else if(c==0)return-1; }if(f)returnRElse return 0;}intMain () {intFlag,i,num,ok,k; while(~scanf("%d%d", n,m)) {if(n==0m==0) Break; V.clear (

HDU 1060 Leftmost Digit (number theory) Leftmost Digit Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission (s): 13680 Accepted Submission (s): 5239 Problem DescriptionGiven a positive integer N, you should output the leftmost digit of N ^ N. InputThe input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case

Problem description Larry graduated this year and finally has a job. He’s making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing problems. The

Modular Multiplication of polynomials Time Limit: 1000MS Memory Limit: 10000K Total submissions: 4377 accepted: 1980 Description consider polynomials whose coefficients are 0 and 1. Addition of two

Link: click here~~ Test instructions Recursion numberProblem DescriptionKnown A (0) = 0, a (1) = 1, a (n) = 3 * A (n-1) + A (n-2) (n≥2) for A (A (A (a (n))) Mod (1e9 + 7)InputThe first line an integer T (t≤10000) represents the number of data

leftmost DigitTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 13680 Accepted Submission (s): 5239Problem Description Given A positive integer N, you should output the leftmost digit of n^n.Input

// This question has never been stuck before, and it takes a night to get through it. // This question is about topological sorting. When there are several error points/topological sorting, m each time, ensure that there is only one entry point of 0,

  Take the logarithm, then obtain the fractional part, and then calculate the power, OK   The Code is as follows: [Cpp]# Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include #

Problem Description Given a positive integer N, you should output the leftmost digit of n^n. Input The input contains several test cases. The ' the ' of the ' input is ' a single integer T which is the number of test cases. T test Cases follow.Each

leftmost Digit Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Problem Description Given A positive integer N, you should output the leftmost digit of n^n.Input the input contains several test cases. The ' the ' of the '

Problem Description Given A positive integer N, you should output the leftmost digit of n^n. Input the input contains several test cases. The ' the ' of the ' input is ' a single integer T which is the number of test cases. T test Cases follow. Each