分类： HDOJ

Problem Description
Given 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.

Input
The first line of the input contains an integer T(1 #include #include #include #include #include #include using namespace std; int n, s, e; int a[100002]; int maxsub(int a[]) { s = e = 1; int sum, j, b, bs, be; sum = b = a[0]; for (j = 0; j= 0 && j != 0) { b = b + a[j]; be = j + 1; } else { b = a[j]; bs = j + 1; be = j + 1; } if (b>sum) { sum = b; s = bs; e = be; } } return sum; } int main() { int t, i, k = 1, sum; scanf(“%d”, &t); while (t–) { scanf(“%d”, &n); for (i = 0; i

另一版本

#include 
using namespace std;
int *arr, first, last, sum;
void MaxSub(int *arr, int count, int &first, int &last, int ∑)
{
	//tFirst,tSum为临时的最大子序列的第一个元素与和
	int tFirst, tSum, i;
	//初始化
	first = tFirst = last = 0;
	sum = tSum = arr[0];
	//遍历该数组
	for (i = 1; i= 0)
			tSum += arr[i];
		else
		{
			tSum = arr[i];
			tFirst = i;
		}
		//如果临时的最大值大于实际的最大值，则更新子序列的first, last, sum
		if (sum > num;
	for (j = 0; j> n;
		arr = new int[n];
		for (i = 0; i> arr[i];
		MaxSub(arr, n, first, last, sum);
		if (top)
			top = false;
		else
			cout 
部分参考自网络。
	

Problem Description
An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. During the rest, it slips down d inches. The process of climbing and resting then repeats. How long before the worm climbs out of the well? We’ll always count a portion of a minute as a whole minute and if the worm just reaches the top of the well at the end of its climbing, we’ll assume the worm makes it out.

Input
There will be multiple problem instances. Each line will contain 3 positive integers n, u and d. These give the values mentioned in the paragraph above. Furthermore, you may assume d

像是小学算术题。没什么好说的，一次过。
放上代码：

#include
int main()
{
	int n, u, d, route, t;
	while (scanf("%d%d%d", &n, &u, &d) && n)
	{
		t = 0;
		route = n;
		while (1)
		{
			route -= u;
			t++;
			if (route 
	

Problem Description
Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In order for him to survive, he decided to create one of the first ciphers. This cipher was so incredibly sound, that no one could figure it out without knowing how it worked.
You are a sub captain of Caesar’s army. It is your job to decipher the messages sent by Caesar and provide to your general. The code is simple. For each letter in a plaintext message, you shift it five places to the right to create the secure message (i.e., if the letter is ‘A’, the cipher text would be ‘F’). Since you are creating plain text out of Caesar’s messages, you will do the opposite:

Cipher text
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Plain text
V W X Y Z A B C D E F G H I J K L M N O P Q R S T U

Only letters are shifted in this cipher. Any non-alphabetical character should remain the same, and all alphabetical characters will be upper case.

Input
Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets. All characters will be uppercase.

A single data set has 3 components:

Start line – A single line, “START”

Cipher message – A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar

End line – A single line, “END”

Following the final data set will be a single line, “ENDOFINPUT”.

Output
For each data set, there will be exactly one line of output. This is the original message by Caesar.

Sample Input
START
NS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJX
END
START
N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ
END
START
IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ
END
ENDOFINPUT

Sample Output
IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSES
I WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROME
DANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE

简单的字符串处理问题，好久没有A题目了，热热身。
注意字符串的判断方式。要在赋值给char后再通过strcmp()函数比较。
读题目，搞清楚到底哪一行是明文，哪一行是密文。
没什么好说的，一次水过。
代码：

#include
#include
#include
int main()
{
	char passwd[201];
	int i, l, cache;
	while (gets(passwd))
	{
		if (strcmp(passwd, "START") == 0)
			continue;
		if (strcmp(passwd, "END") == 0)
			continue;
		if (strcmp(passwd, "ENDOFINPUT") == 0)
			break;
		l = strlen(passwd);
		for (i = 0; i = 70 && passwd[i] = 65)
			{
				passwd[i] += 21;
			}
		}
		for (i = 0; i 
	

Problem Description
A positive integer x can represent as (a1a2…akak…a2a1)10 or (a1a2…ak−1akak−1…a2a1)10 of a 10-based notational system, we always call x is a Palindrome Number. If it satisfies 0<a1<a2<…<ak≤9, we call x is a Beautiful Palindrome Number.
Now, we want to know how many Beautiful Palindrome Numbers are between 1 and 10N.

Input
The first line in the input file is an integer T(1≤T≤7), indicating the number of test cases.
Then T lines follow, each line represent an integer N(0≤N≤6).

Output
For each test case, output the number of Beautiful Palindrome Number.

Sample Input
2
1
6

Sample Output
9
258

要求判断范围内有多少回文串。由于范围小，可以直接打表计算。

#include
int main()
{
	int a[7]={1,9,18,54,90,174,258},t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		printf("%dn",a[n]);
	}
return 0;
}

#include
#include
using namespace std;
char a[100000], b[100000];
bool is(char *p)
{
	for (; *p != ' '; p++)
	if (*p == '.')return true;
	return false;
}

void det(char *p)
{
	for (; *p != ' '; p++);
	p--;
	for (; *p == '0'; p--)
		*p = ' ';
	if (*p == '.')*p = ' ';
}

int main()
{
	while (cin >> a >> b)
	{
		if (is(a))det(a);
		if (is(b))det(b);
		if (strcmp(a, b) == 0)cout 
	

Input

Output

Sample Input

Sample Output

#include
#include
#include 
#include
#include
using namespace std;
double ai[100];
int main(){
	int n, i, j, cas;
	double res;
	scanf("%d",&cas);
	while (cas--){
		scanf("%d", &n);
		res = 0;
		for (i = 0; i= 0; i--){
			res += pow(0.95, n - 1 - i)*ai[i];
		}
		printf("%.10lfn", res);
	}
	return 0;
}

Problem Description
A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n – 1) + B * f(n – 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 #include int main(){     long a,b,n,i,func[100];     while(scanf(“%ld%ld%ld”,&a,&b,&n) && a!=0){         func[1]=func[2]=1;         for(i=3;i48)?n%48:n]);     }     return 0; }

Problem Description 输入n(n<100)个数，找出其中最小的数，将它与最前面的数交换后输出这些数。

Input 输入数据有多组，每组占一行，每行的开始是一个整数n，表示这个测试实例的数值的个数，跟着就是n个整数。n=0表示输入的结束，不做处理。

Output 对于每组输入数据，输出交换后的数列，每组输出占一行。

Sample Input

4 2 1 3 4

5 5 4 3 2 1

0

Sample Output

1 2 3 4

1 4 3 2 5

#include
#include
int main()
{
	int n, ni;
	int num[100];
	int a, cache,ci,min;
	while (~scanf("%d", &n)&&n!=0)
	{
		memset(num, 0, sizeof(num));
		for (ni = 0; ni