1.1 How to use USACO

文件流
c

c

限制

题型

1_1_2 ride 字符串处理

1_1_5分糖

1_1_6黑色星期五

1_1_7 deads

5/23更新:

在这个网站找到正确使用xcode方法

采用的排名第二的step by step 方法

Use the Product Menu(软件打开上面那一条)

Select the Scheme option(下面就直接找就行了…)

Select the Edit Scheme option

Click the Options Tab

Tick the “Working Directory” item

Click the small icon at the right end of the text box.

Select the directory

文件流

c

/*
ID: leidar11
LANG: C
TASK: test
*/
#include <stdio.h>
main () {
FILE *fin  = fopen ("test.in", "r");
FILE *fout = fopen ("test.out", "w");
int a, b;
fscanf (fin, "%d %d", &a, &b);  /* the two input integers */
fprintf (fout, "%d\n", a+b);
exit (0);
}

实际xcode路径

FILE *fin = fopen("/Users/macbook/Desktop/in.txt", "r");
FILE *fout = fopen("/Users/macbook/Desktop/out.txt", "w");

c + +

Below is a simple solution in the C++ programming language. Note the use of 'return (0);', which is usually required to exit properly.

/*
ID: leidar11
PROG: test
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>

using namespace std;

int main() {
ofstream fout ("test.out");
ifstream fin ("test.in");
int a, b;
fin >> a >> b;
fout << a+b << endl;
return 0;
}

实际Xcode路径：

ofstream fout ("/Users/macbook/Desktop/out.txt");
ifstream fin ("/Users/macbook/Desktop/in.txt");

另一种方法

string inputFilename = "beads.in", outputFilename = "beads.out";
input.open(inputFilename.c_str(), ios::in);
output.open(outputFilename.c_str(), ios::out);

限制

The restrictions are few:

One second runtime limit unless other specified (700 MHz Pentium III)

About 16MB datasize limit

About 1MB stacksize limit

In C/C++, ints are 32 bits (char is 8; short is 16; long is 32; long long is 64)

Be sure your program exits with status 0

Be sure you print complete lines (with terminating newline), - - not just a few words or numbers

Don’t use files other than the specified input, output, and auxiliary files

Other common sense rules that need not be listed

题型

Dynamic Programming

Knapsack背包

Greedy

search

Complete Search

Flood Fill

Shortest Path

Recursive Search Techniques

Heuristic Search启发式

Approximate Search

Minimum Spanning Tree

Computational Geometry

Network Flow

Eulerian Path

Two-Dimensional Convex Hull

BigNums

Ad Hoc Problems 临时问题

1_1_2 ride 字符串处理

analysis_hash

int hash(char *s)
{
int i, h;

h = 1;
for(i=0; s[i] && isalpha(s[i]); i++)
h = ((s[i]-'A'+1)*h) % 47;
return h;
}

1_1_5分糖

字符串处理比较

#include <string.h>
#include <assert.h>

void addperson(char *name){
assert(npeople < MAXPEOPLE);
strcpy(people[npeople].name, name);
npeople++;
}

Person* lookup(char *name){
int i;
/* look for name in people table */
for(i=0; i<npeople; i++)
if(strcmp(name, people[i].name) == 0)
return &people[i];
assert(0);  /* should have found name */
}

map水过..BUT

gift写成gist RE两次

1_1_6黑色星期五

判断闰年 暴力枚举

Brute force

int isleap(int y){
return y%4==0 && (y%100 != 0 || y%400 == 0);
}

int mtab[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

/* return length of month m in year y */
int mlen(int y, int m){
if(m == 1)    /* february */
return mtab[m]+isleap(y);
else
return mtab[m];
}

1_1_7 deads

神烦循环字符串…

题解都复制了一遍:)

题解二算一边= =

Daniel Bundala had an O(n) solution:

Dynamic Programming is good method for solving this problem in O(N). If we consider two copies of the string we easy transform cyclic configuration of the necklace to linear. Now we can compute for each breaking point how many beads of the same color can be collected on the left and on the right from the breaking point. I show how we can compute it only for the left side. For right side it is analogical. Let r[p] and b[p] be the number of red / blue beads that can be collected, when necklace is broken in point p. If we know this and color of next bead (c) we can compute r[p+1] and b[p+1].
r[0] = p[0] = 0
If c = 'r' then r[p+1] = r[p] + 1 and b[p+1] = 0
because the length of the blue beads is 0.
if c = 'b' then b[p+1] = b[p] + 1 and r[p+1] = 0
if c = 'w' then both length of the red and length of blue beads
can be longer.
so r[p+1] = r[p]+1 and b[p+1] = b[p] + 1.
The number of beads that can be collected in breaking point p is then max(left[r[p]], left[b[p]]) + max(right[r[p]], right[b[p]]). And the maximum from this value is answer for the problem.
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

FILE *in,*out;

int main () {
in = fopen("beads.in", "r");
out = fopen ("beads.out", "w");
int n;
char tmp[400], s[800];
fscanf(in, "%d %s", &n, tmp);

strcpy(s, tmp);
strcat(s, tmp);

int left[800][2], right[800][2];
left[0][0] = left[0][1] = 0;

for (int i=1; i<= 2 * n; i++){
if (s[i - 1] == 'r'){
left[i][0] = left[i - 1][0] + 1;
left[i][1] = 0;
} else if (s[i - 1] == 'b'){
left[i][1] = left[i - 1][1] + 1;
left[i][0] = 0;
} else {
left[i][0] = left[i - 1][0] + 1;
left[i][1] = left[i - 1][1] + 1;
}
}

right[2 * n][0] = right[2 * n][1] = 0;
for (int i=2 * n - 1; i >= 0; i--){
if (s[i] == 'r'){
right[i][0] = right[i + 1][0] + 1;
right[i][1] = 0;
} else if (s[i] == 'b'){
right[i][1] = right[i + 1][1] + 1;
right[i][0] = 0;
} else {
right[i][0] = right[i + 1][0] + 1;
right[i][1] = right[i + 1][1] + 1;
}
}

int m = 0;
for (int i=0; i<2 * n; i++)
m = max(m, max(left[i][0], left[i][1]) + max(right[i][0], right[i][1]));
m = min(m, n);
fprintf(out, "%d\n", m);
fclose(in); fclose(out);
return 0;
}

题解二:

for(i=0; i<n; i++) {
c = (char) s[i];
if(c == 'w')
state = 0;
else
state = 1;
j = i;
current = 0;

while(state <= 2) {
// dont go further in second string than starting position in first string
while(j<n+i && (s[j] == c || s[j] == 'w')) {
current++;
j++;
}
state++;
c = s[j];
} // while
if(current > max)
max = current;
} // for