杭电-acm

2012/5/9 LEEREVWARE HDUACM[键入文档副标题] | AdministratorMax Sum Plus Plus(1024)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9131 Accepted Submission(s): 2993Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.Given a consecutive number sequence S1, S2, S3, S4 ... S x, ... S n(1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x≤ 32767). We define a function sum(i, j) = S i + ... + S j(1 ≤ i ≤ j ≤ n).Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(i m, j m) maximal (i x≤ i y≤ j x or i x≤ j y≤ j x is not allowed).But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^InputEach test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... S n. Process to the end of file.OutputOutput the maximal summation described above in one line.Sample Input1 3 12 32 6 -1 4 -23 -2 3N!(1042)Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 31255 Accepted Submission(s): 8631Problem DescriptionGiven an integer N(0 ≤ N ≤ 10000), your task is to calculate N!InputOne N in one line, process to the end of file.OutputFor each N, output N! in one line.Sample Input123Sample Output126#include <stdio.h>int main(){int i,j,n,u,arry;while(scanf("%d",&n)!=EOF){int a[10000]={1};u=0;for(i=1;i<=n;i++){arry=0;for(j=0;j<=u;j++){arry+=a[j]*i;a[j]=arry%10000;arry/=10000;}if(arry){u++;a[u]=arry;}}printf("%d",a[u]);for(i=u-1;i>=0;i--)printf("%04d",a[i]);printf("\n");}return 0;}Max Sum(1003)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 75802 Accepted Submission(s): 17388Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.InputThe first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).OutputFor each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.Sample Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5Sample OutputCase 1:14 1 4Case 2:7 1 6#include <stdio.h>int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);int i,j,t,n,m,x1,x,y,max,a;scanf("%d",&t);for(i=1;i<=t;i++){scanf("%d",&n);m=0;x1=0;x=0;y=0;max=-1000;for(j=0;j<n;j++){scanf("%d",&a);if(m<0){m=a;x1=j;}elsem+=a;if(max<m){max=m;x=x1;y=j;}}printf("Case %d:\n%d %d %d\n",i,max,x+1,y+1);}return 0;}Let the Balloon Rise(1004)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40420 Accepted Submission(s): 14044Problem DescriptionContest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.This year, they decide to leave this lovely job to you.InputInput contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.A test case with N = 0 terminates the input and this test case is not to be processed.OutputFor each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.Sample Input5greenredblueredred3pinkorangepinkSample Outputredpink#include <stdio.h>#include<string.h>int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);int i,j,k,n,l,max;char str[1000][15];while(scanf("%d",&n)!=EOF&&n){getchar();int a[1000]={0};char ch[1000][15]={""};for(i=0;i<n;i++)scanf("%s",str[i]);strcpy(ch[0],str[0]);j=0;a[0]=1;for(i=1;i<n;i++){l=0;for(k=0;k<=j;k++)if(strcmp(str[i],ch[k])==0){l=1;a[k]++;break;}if(l==0){j++;strcpy(ch[j],str[i]);a[j]++;}}max=0;for(i=0;i<=j;i++)if(max<a[i]){max=a[i];k=i;}printf("%s\n",ch[k]);}return 0; }Tick and Tick(1006)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4797 Accepted Submission(s): 1284Problem DescriptionThe three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.InputThe input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.OutputFor each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.Sample Input12090-1Sample Output100.0000.0006.251严重超时（需改进算法）#include <stdio.h>double g(double a,double b){while(a>360)a-=360;while(b>360)b-=360;double min;if(a<b)min=(b-a<a-b+360?b-a:a-b+360);elsemin=(a-b<b-a+360?a-b:b-a+360);return min;}double f(double a,double b,double c){double min=g(a,b);min=(min<g(b,c)?min:g(b,c));min=(min<g(a,c)?min:g(a,c));return min;}int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);double t,s,m,h,a,num;double p;while(scanf("%lf",&a)!=EOF&&a!=-1){num=0;for(t=0;t<43200;t+=0.01){s=t*6;m=t/10.0;h=t/120.0;if(f(s,m,h)>=a)num+=0.01;}p=num/432.0;printf("%.3lf\n",p);}return 0; }改进后#include <stdio.h>double g(double a,double b){double min;if(a<b)min=(b-a<a-b+360?b-a:a-b+360);elsemin=(a-b<b-a+360?a-b:b-a+360);return min;}double f(double a,double b,double c){double min=g(a,b);min=(min<g(b,c)?min:g(b,c));min=(min<g(a,c)?min:g(a,c));return min;}int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);double t,s,m,h,a,num;double p;while(scanf("%lf",&a)!=EOF&&a!=-1){num=0;s=0;m=0;h=0;for(t=0;t<43200;t+=0.1,s+=0.6,m+=0.01,h+=0.000833333) {if(s>360)s-=360;if(m>360)m-=360;if(f(s,m,h)>=a)num+=0.1;}p=num/432.0;printf("%.3lf\n",p);}return 0;}钥匙计数之二(1480)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 692 Accepted Submission(s): 419Problem Description一把钥匙有N个槽，2<N<26槽深为1，2，3，4,5,6。