Cool J has recently become a businessman Mr. Jackson, and he has to make a lot of phone calls now. Today he has n calls planned. For each call we know the moment ti (in seconds since the start of the day) when it is scheduled to start and its duration di (in seconds). All ti are different. Mr. Jackson is a very important person, so he never dials anybody himself, all calls will be incoming.
Mr. Jackson isn't Caesar and he can't do several things at once. If somebody calls him while he hasn't finished the previous conversation, Mr. Jackson puts the new call on hold in the queue. In this case immediately after the end of the current call Mr. Jackson takes the earliest incoming call from the queue and starts the conversation. If Mr. Jackson started the call at the second t, and the call continues for d seconds, then Mr. Jackson is busy at seconds t, t + 1, ..., t + d - 1, and he can start a new call at second t + d. Note that if Mr. Jackson is not busy talking when somebody calls, he can't put this call on hold.
Mr. Jackson isn't Napoleon either, he likes to sleep. So sometimes he allows himself the luxury of ignoring a call, as if it never was scheduled. He can ignore at most k calls. Note that a call which comes while he is busy talking can be ignored as well.
What is the maximum number of seconds Mr. Jackson can sleep today, assuming that he can choose an arbitrary continuous time segment from the current day (that is, with seconds from the 1-st to the 86400-th, inclusive) when he is not busy talking?
Note that some calls can be continued or postponed to the next day or even later. However, the interval for sleep should be completely within the current day.
Output
Print a number from 0 to 86400, inclusive — the maximally possible number of seconds for Mr. Jackson to sleep today.
Note
In the first sample the most convenient way is to ignore the first two calls.
In the second sample it is best to ignore the third call. In this case Mr. Jackson will have been speaking:
- first call: from 1-st to 20000-th second,
- second call: from 20001-st to 30000-th second,
- fourth call: from 30001-st to 40000-th second (the third call is ignored),
- fifth call: from 80000-th to 139999-th second.
Thus, the longest period of free time is from the 40001-th to the 79999-th second.
DP；；；
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int dp[4005][4005];
struct node
{
int s,d;
};
node a[4005];
int main()
{
int i,j,n,k,ans;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=0;i<n;++i)
{
scanf("%d%d",&a[i].s,&a[i].d);
}
ans=0;
for(i=0;i<=n;++i)
{
for(j=0;j<=k;++j)
{
dp[i][j]=86401;
}
}
dp[0][0]=1;
for(i=0;i<n;++i)
{
for(j=0;j<=k;++j)
{
if(j!=k)
dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]);
if(dp[i][j]<a[i].s)
{
ans=max(ans,a[i].s-dp[i][j]);
dp[i+1][j]=min(dp[i+1][j],a[i].s+a[i].d);
}
else
{
dp[i+1][j]=min(dp[i+1][j],dp[i][j]+a[i].d);
}
}
}
for(j=0;j<=k;++j)
{
ans=max(ans,86401-dp[n][j]);
}
printf("%d\n",ans);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/05/2380629.html
The Berland University is preparing to celebrate the 256-th anniversary of its founding! A specially appointed Vice Rector for the celebration prepares to decorate the campus. In the center of the campus n ice sculptures were erected. The sculptures are arranged in a circle at equal distances from each other, so they form a regular n-gon. They are numbered in clockwise order with numbers from 1 to n.
The site of the University has already conducted a voting that estimated each sculpture's characteristic of ti — the degree of the sculpture's attractiveness. The values of ti can be positive, negative or zero.
When the university rector came to evaluate the work, he said that this might be not the perfect arrangement. He suggested to melt some of the sculptures so that:
- the remaining sculptures form a regular polygon (the number of vertices should be between 3 and n),
- the sum of the ti values of the remaining sculptures is maximized.
Help the Vice Rector to analyze the criticism — find the maximum value of ti sum which can be obtained in this way. It is allowed not to melt any sculptures at all. The sculptures can not be moved.
Output
Print the required maximum sum of the sculptures' attractiveness.
Note
In the first sample it is best to leave every second sculpture, that is, leave sculptures with attractivenesses: 2, 4, 5 и 3.
暴力求解的。。。
就是等間隔取數(shù)，求其和最大值！！！
#include<stdio.h>
int aa[20110];
int main()
{
int n;
int res;
int sum;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&aa[i]);
res=-200000000;
for(int i=1;i<=n/3;i++)
{
if(n%i!=0)continue;
if(n/i<3)continue;
for(int j=1;j<=i;j++)
{
sum=0;
for(int k=j;k<=n;k+=i)
sum+=aa[k];
if(sum>res) res=sum;
}
}
printf("%d\n",res);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/05/2380627.html
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Note
In the first test we can sort the children into four cars like this:
- the third group (consisting of four children),
- the fourth group (consisting of three children),
- the fifth group (consisting of three children),
- the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
水題，4個(gè)人的單獨(dú)。
3和1搭配，
2和2搭配。
3有多的話只能單獨(dú)了。
2可以加入1個(gè)的
#include<stdio.h>
int aa[5];
int main()
{
int n;
int t;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<5;i++)aa[i]=0;
while(n--)
{
scanf("%d",&t);
aa[t]++;
}
int res=0;
res+=aa[4];
if(aa[1]==aa[3])
{
res+=aa[1];
aa[1]=0;
aa[3]=0;
if(aa[2]%2==0)
{
res+=aa[2]/2;
}
else
{
res=res+aa[2]/2+1;
}
}
else if(aa[1]<aa[3])
{
res+=aa[1];
aa[3]-=aa[1];
aa[1]=0;
res+=aa[3];
if(aa[2]%2==0)
{
res+=aa[2]/2;
}
else
{
res=res+aa[2]/2+1;
}
}
else
{
res+=aa[3];
aa[1]-=aa[3];
aa[3]=0;
if(aa[2]%2==0)
{
res+=aa[2]/2;
}
else
{
res=res+aa[2]/2+1;
aa[1]-=2;
}
if(aa[1]>=1)
{
if(aa[1]%4==0) res+=aa[1]/4;
else
res=res+aa[1]/4+1;
}
}
printf("%d\n",res);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/05/2380624.html
"Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round.
Output
Output the number of participants who advance to the next round.
Note
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
很簡(jiǎn)單的題目~~~~~
#include<stdio.h>
int mm[120];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&mm[i]);
int res=k;
if(mm[k]>0)
{
for(int i=k+1;i<=n;i++)
{
if(mm[i]<mm[k])break;
res++;
}
}
else
{
res--;
for(int i=k-1;i>=1;i--)
{
if(mm[i]>0) break;
res--;
}
}
printf("%d\n",res);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/05/2380621.html
Problem E
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Aunt Lizzie takes half a pill of a certain medicine every day. She starts with a bottle that contains N pills.
On the first day, she removes a random pill, breaks it in two halves, takes one half and puts the other half back into the bottle.
On subsequent days, she removes a random piece (which can be either a whole pill or half a pill) from the bottle. If it is half a pill, she takes it. If it is a whole pill, she takes one half and puts the other half back into the bottle.
In how many ways can she empty the bottle? We represent the sequence of pills removed from the bottle in the course of 2N days as a string, where the i-th character is W if a whole pill was chosen on the i-th day, and H if a half pill was chosen (0 <= i < 2N). How many different valid strings are there that empty the bottle?
Input
The input will contain data for at most 1000 problem instances. For each problem instance there will be one line of input: a positive integer N <= 30, the number of pills initially in the bottle. End of input will be indicated by 0.
Output
For each problem instance, the output will be a single number, displayed at the beginning of a new line. It will be the number of different ways the bottle can be emptied.
Sample Input
Sample Output
132 1 14 2 5 3814986502092304
相當(dāng)于求N個(gè)W，和N個(gè)H的排列數(shù)，而且要求前面任意個(gè)中必需H的個(gè)數(shù)不大于W的個(gè)數(shù)~~~~
不知道哪些大神是怎么樣做出來(lái)的，好快的速度就做出來(lái)了~~~~
我想了好久都沒(méi)有想到DP的方法，
最后想出來(lái)了，感覺(jué)方法比較笨~~~~
轉(zhuǎn)化成了圖，W表示往下走，H表是往右走，
則必須在左下角中。
N個(gè)W和H的時(shí)候就是到N-1行的路徑數(shù)
如N=1時(shí)，為dp[0,0]
N=2時(shí)為dp[1,0]+dp[1,1]
N=3時(shí)為dp[2,0]+dp[2,1]+dp[2,2]
應(yīng)該會(huì)有更好的方法的~~~~~
我的思路有點(diǎn)蹉了~~~~但幸好做出來(lái)了！
代碼如下：
#include<stdio.h>
long long dp[31][31];
void init()
{
dp[0][0]=1;
for(int i=1;i<31;i++)
{
dp[i][0]=1;
for(int j=1;j<i;j++)
dp[i][j]=dp[i-1][j]+dp[i][j-1];
dp[i][i]=dp[i][i-1];
}
}
int main()
{
int N;
init();
while(scanf("%d",&N),N)
{
printf("%I64d\n",dp[N][N]);
}
return 0;
}
要求的其實(shí)就是從（0，0）到(N,N)的路徑數(shù)，只能往下或者往右走，而且不能走到右上部分去。 文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/03/2378248.html
Problem A
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
You are given a list of N non-negative integers a(1), a(2), ... , a(N). You replace the given list by a new list: the k-th entry of the new list is the absolute value of a(k) - a(k+1), wrapping around at the end of the list (the k-th entry of the new list is the absolute value of a(N) - a(1)). How many iterations of this replacement are needed to arrive at a list in which every entry is the same integer?
For example, let N = 4 and start with the list (0 2 5 11). The successive iterations are:
2 3 6 11
1 3 5 9
2 2 4 8
0 2 4 6
2 2 2 6
0 0 4 4
0 4 0 4
4 4 4 4
Thus, 8 iterations are needed in this example.
Input
The input will contain data for a number of test cases. For each case, there will be two lines of input. The first line will contain the integer N (2 <= N <= 20), the number of entries in the list. The second line will contain the list of integers, separated by one blank space. End of input will be indicated by N = 0.
Output
For each case, there will be one line of output, specifying the case number and the number of iterations, in the format shown in the sample output. If the list does not attain the desired form after 1000 iterations, print 'not attained'.
Sample Input
4 0 2 5 11 5 0 2 5 11 3 4 300 8600 9000 4000 16 12 20 3 7 8 10 44 50 12 200 300 7 8 10 44 50 3 1 1 1 4 0 4 0 4 0
Sample Output
Case 1: 8 iterations Case 2: not attained Case 3: 3 iterations Case 4: 50 iterations Case 5: 0 iterations Case 6: 1 iterations
水題~~~~~
迭代就可以了！
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int a[50];
int main()
{
int N,res,j;
int iCase=0;
while(scanf("%d",&N),N)
{
iCase++;
for(int i=0;i<N;i++)
{
scanf("%d",&a[i]);
}
for(j=0;j<N-1;j++)
{
if(a[j]!=a[j+1])break;
}
if(j>=N-1)
{
printf("Case %d: 0 iterations\n",iCase);
continue;
}
for(res=1;res<=1000;res++)
{
a[N]=a[0];
for(int i=0;i<N;i++)
{
a[i]=abs(a[i]-a[i+1]);
}
for(j=0;j<N-1;j++)
{
if(a[j]!=a[j+1])break;
}
if(j>=N-1)break;
}
if(res<=1000)printf("Case %d: %d iterations\n",iCase,res);
else printf("Case %d: not attained\n",iCase);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/03/03/2378235.html
| 店長(zhǎng)推薦Ⅱ |
| Time Limit: 1000 MS |
Memory Limit: 65536 K |
| Total Submit: 97(31 users) |
Total Accepted: 28(24 users) |
Special Judge: No |
|
| Description |
|
想必大家已經(jīng)領(lǐng)略了店長(zhǎng)的調(diào)皮,自從店長(zhǎng)知道了vagaa之后就一直沉迷期中不能自拔,教主看到后很是傷心，玩物喪志啊!于是教主教店長(zhǎng)了一個(gè)vagaa的新用法,資源搜索功能。輸入一些關(guān)鍵詞后可以搜索出相關(guān)共享的好資源，店長(zhǎng)得知后又是欣喜若狂。同時(shí),教主又發(fā)明一個(gè)游戲，是上次的升級(jí)版,這次給出一些由字母和數(shù)字的玩具的同時(shí)，關(guān)鍵字不再是vagaa了，需要自己給出.看最后能組成多少個(gè)關(guān)鍵字。
|
| Input |
|
每行輸入一個(gè)字符串由小寫(xiě)字母和數(shù)字組成，每個(gè)字符代表一個(gè)玩具, 緊接著輸入一個(gè)關(guān)鍵字，同樣由字母和數(shù)字組組成，字符串長(zhǎng)度0<n<=10000，關(guān)鍵字長(zhǎng)度0<m<=200
處理到文件結(jié)束
|
| Output |
|
輸出能夠組成關(guān)鍵字的數(shù)量
按照樣例輸出格式輸出并換行.
|
| Sample Input |
|
vagaadsfaagav ga
asgvasdfzxcades dea
|
| Sample Output |
|
Case 1: 2
Case 2: 1
|
| Author |
| void |
#include<stdio.h>
#include<string.h>
int tt[40],count[40];
char str[10010];
int main()
{
int len,m;
int iCase=0;
while(scanf("%s",&str)!=EOF)
{
iCase++;
len=strlen(str);
memset(tt,0,sizeof(tt));
memset(count,0,sizeof(count));
for(int i=0;i<len;i++)
{
if(str[i]>='0'&&str[i]<='9')
{
m=str[i]-'0'+0;
tt[m]++;
}
else
{
m=str[i]-'a'+10;
tt[m]++;
}
}
scanf("%s",&str);
len=strlen(str);
for(int i=0;i<len;i++)
{
if(str[i]>='0'&&str[i]<='9')
{
m=str[i]-'0'+0;
count[m]++;
}
else
{
m=str[i]-'a'+10;
count[m]++;
}
}
int res=10000;
for(int i=0;i<=9+26;i++)
{
if(count[i]>0)
{
int tmp=tt[i]/count[i];
if(tmp<res) res=tmp;
}
}
printf("Case %d: %d\n",iCase,res);
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/02/27/2370325.html
| 青蛙過(guò)河 |
| Time Limit: 1000 MS |
Memory Limit: 65535 K |
| Total Submit: 29(11 users) |
Total Accepted: 9(9 users) |
Special Judge: No |
|
| Description |
在河上有一座獨(dú)木橋，一只青蛙想沿著獨(dú)木橋從河的一側(cè)跳到另一側(cè)。在橋上有一些石子，青蛙很討厭踩在這些石子上。由于橋的長(zhǎng)度和青蛙一次跳過(guò)的距離都是正整數(shù)，我們可以把獨(dú)木橋上青蛙可能到達(dá)的點(diǎn)看成數(shù)軸上的一串整點(diǎn)：0，1，……，L（其中L是橋的長(zhǎng)度）。坐標(biāo)為0的點(diǎn)表示橋的起點(diǎn)，坐標(biāo)為L(zhǎng)的點(diǎn)表示橋的終點(diǎn)。青蛙從橋的起點(diǎn)開(kāi)始，不停的向終點(diǎn)方向跳躍。一次跳躍的距離是s到t之間的任意正整數(shù)（包括s,t）。當(dāng)青蛙跳到或跳過(guò)坐標(biāo)為L(zhǎng)的點(diǎn)時(shí)，就算青蛙已經(jīng)跳出了獨(dú)木橋。
題目給出獨(dú)木橋的長(zhǎng)度L，青蛙跳躍的距離范圍s,t，橋上石子的位置。你的任務(wù)是確定青蛙要想過(guò)河，最少需要踩到的石子數(shù)。 |
| Input |
有多組測(cè)試數(shù)據(jù)。
對(duì)于每組測(cè)試數(shù)據(jù)，第一行四個(gè)正整數(shù)L, s, t, n（1 <= L <= 10^5, 1 <= s <= t <= 10，1 <= n <= 100），分別表示獨(dú)木橋的長(zhǎng)度，青蛙一次跳躍的最小距離，最大距離，及橋上石子的個(gè)數(shù)。第二行有n個(gè)不同的正整數(shù)分別表示這n個(gè)石子在數(shù)軸上的位置（數(shù)據(jù)保證橋的起點(diǎn)和終點(diǎn)處沒(méi)有石子）。所有相鄰的整數(shù)之間用一個(gè)空格隔開(kāi)。 |
| Output |
| 每組測(cè)試數(shù)據(jù)僅輸出一行，包括一個(gè)整數(shù)，表示青蛙過(guò)河最少需要踩到的石子數(shù)。 |
| Sample Input |
10 2 3 5
2 3 5 6 7 |
| Sample Output |
|
2
#include<stdio.h>
#include<string.h>
const int MAXN=100020;
int flag[MAXN];
int dp[MAXN];
int main()
{
int L,s,t,n;
int a;
while(scanf("%d%d%d%d",&L,&s,&t,&n)!=EOF)
{
memset(flag,0,sizeof(flag));
memset(dp,-1,sizeof(dp));//初始化，-1為不能到達(dá)的
//dp[i]表示到底 i 點(diǎn)需要經(jīng)過(guò)的最少石子數(shù)，-1表示不能到達(dá)
for(int i=0;i<n;i++)
{
scanf("%d",&a);
flag[a]=1;//有石子為1，否則為0
}
dp[0]=0;
for(int i=s;i<=L+t-1;i++)
{
for(int j=i-t;j<=i-s;j++)// j 點(diǎn)跳到 i 點(diǎn)
{
if(j>=0&&dp[j]!=-1)//j 點(diǎn)能夠跳到
{
if(dp[i]==-1)dp[i]=dp[j]+flag[i]; //第一次 直 接 給 值
else if(dp[i]>dp[j]+flag[i]) dp[i]=dp[j]+flag[i];//找小的值
}
}
}
int res=10000;
for(int i=L;i<=L+t-1;i++)//L 到 L+t-1 中最小的非 -1 值
{
if(dp[i]!=-1&&dp[i]<res) res=dp[i];
}
printf("%d\n",res);
}
return 0;
}
|
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/02/27/2369901.html
2011已經(jīng)逝去，2012已經(jīng)到來(lái)~~~~
2011年，總感覺(jué)自己得到了很多，也失去了很多，也留下不盡的遺憾。仍記得2011年到來(lái)的時(shí)候，我自己對(duì)自己說(shuō)這一年將會(huì)是我關(guān)鍵的一年，給自己定下了很多目標(biāo)~~~~這些目標(biāo)只能說(shuō)是大致實(shí)現(xiàn)了吧，也很多遺憾！
為了迎接2012年，總感覺(jué)自己應(yīng)該寫(xiě)點(diǎn)東西，來(lái)總結(jié)自己的2011，規(guī)劃下2012.但是一直抽不出時(shí)間來(lái)寫(xiě)。等有時(shí)間了再寫(xiě)一篇來(lái)總結(jié)2011吧！
人生應(yīng)該處于不斷的奮斗當(dāng)中！2012年，我要取得成功！2012年，我一定會(huì)做得更好！
加油！
————————————————————————————————————by kuangbin 文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2012/01/03/2310575.html
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1006
這題做了好久了，學(xué)習(xí)了網(wǎng)上的解法后自己寫(xiě)出來(lái)了。
就是枚舉12*60分鐘，看一分鐘內(nèi)有多少秒是happytime，一分鐘內(nèi)解三個(gè)不等式得到區(qū)間。
具體看代碼。
代碼有注釋很詳細(xì)的。
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct qujian
{
double l,r;
};
double D;
qujian solve(double a,double b)//解方程 D<=a*x+b<=360-D ,并且和 [0,60] 取交集
{
qujian p;
if(a>0)
{
p.l=(D-b)/a;
p.r=(360-D-b)/a;
}
else
{
p.l=(360-D-b)/a;
p.r=(D-b)/a;
}
if(p.l<0)p.l=0;
if(p.r>60)p.r=60;
if(p.l>=p.r) p.l=p.r=0;
return p;
}
qujian jiao(qujian a,qujian b)
{
qujian p;
p.l=max(a.l,b.l);
p.r=min(a.r,b.r);
if(p.l>=p.r) p.l=p.r=0;
return p;
}
/*
hh=30*h+m/2+s/120;
mm=6*m+s/10;
ss=6*s;
D<=|hh-mm|<=360-D;
D<=|hh-ss|<=360-D;
D<=|mm-ss|<=360-D;
hh-mm=
*/
double happytime(int h,int m)//計(jì)算 h 時(shí)，m 分 滿足題意的秒數(shù)
{
double a,b;
qujian s[3][2];
qujian s1;
//解方程 D<=|hh-mm|<=360-D
//hh=30*h+m/2+s/120;
//mm=6*m+s/10;
a=1.0/120-0.1;
b=30*h+m/2.0-6*m;
s[0][0]=solve(a,b);
s[0][1]=solve(-a,-b);
//解方程 D<=|hh-ss|<=360-D
//hh=30*h+m/2+s/120;
//ss=6*s;
a=1.0/120-6.0;
b=30*h+m/2.0;
s[1][0]=solve(a,b);
s[1][1]=solve(-a,-b);
//解方程 D<=|mm-ss|<=360-D
//mm=6*m+s/10;
//ss=6*s;
a=0.1-6;
b=6*m;
s[2][0]=solve(a,b);
s[2][1]=solve(-a,-b);
double res=0;
//六個(gè)區(qū)間，選三個(gè)取交集。
//因?yàn)榻^對(duì)值的式子得到的兩個(gè)區(qū)間要并，而三個(gè)不同表達(dá)式的區(qū)間要交，故這樣做。
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
{
s1=jiao(jiao(s[0][i],s[1][j]),s[2][k]);
res+=s1.r-s1.l;
}
return res;
}
int main()
{
while(scanf("%lf",&D))
{
if(D==-1) break;
double res=0;
int h,m;
for(h=0;h<12;h++)
{
for(m=0;m<60;m++)
{
res+=happytime(h,m);
}
}
printf("%.3lf\n",res*100.0/(60*60*12));
}
return 0;
}
文章來(lái)源: http://www.cnblogs.com/kuangbin/archive/2011/12/04/2275470.html
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