Getting algorithm contest (a) language papers loop structure

Master list:

for while do..while 循环
Counter and accumulator
The method of debugging intermediate results output
Timing functions with efficiency test program
Redirection methods used to read and write files
Reading and writing files with fopen
Construction of the local operating environment with conditions pragma
-Wall compiler option with more warnings

A, for circulation

Program. 1: for loop output

#include<stdio.h>
int main(){
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
        printf("%d\n",i);
    return 0;
}

Details: In the loop variable i is defined, and therefore not visible in vitro cycle i

Detailed explanation

# Suggestions proposed to shorten the range of the variables defined ------- e.g., initialization portion in a for loop, the loop variable is defined

Procedure 2: 4 perfect square outputs of all forms of aabb

#include <stdio.h> 
#include <math.h>
 // AABB 4 digits, so the range a --1-9, range b --0-9 
int main () {
     for ( int a = . 1 ; a <= . 9 ; a ++ ) {
         for ( int B = 0 ; B <= . 9 ; B ++ ) {
             // start determination, aabb is not the number of square 
            int n-= a * 1100 is + B * . 11 ;
             // with a memory variable m sqrt (n) after the rounded integer, then it is determined whether m is equal to m * n-
             // Floor (x) function returns the largest integer not exceeding x 
            int m = Floor (sqrt (n) + 0.5);
            if(m*m == n)
                printf("%d\n",n);
        }
    }
    return 0;
}

Results: 7744

Floor (sqrt (n-) + 0.5 )   represents rounding

# Floating point computation errors may exist, it is assumed after a lot of calculation, due to the influence of the error, becomes an integer of 1 0.99999999, floor of the result will be 0 instead of 1, in order to reduce the influence of the error is generally rounded to , i.e., floor (x + 0.5)

# Floating point error may exist, performing floating-point comparison, floating point error should be taken into

Another approach is to enumerate the square root of x, so as to avoid square root operations

#include<stdio.h>
int main(){
    for(int x = 1; ; x++){
        int n = x *x;
        if (n < 1000)
            continue;
        if (n > 9999)
            break;
        int hi = n/100;
        int lo = n%100;
        if( hi/10 == hi%10 && lo/10 == lo%10){
            printf("%d\n",n);
        }
    }
    return 0;
}

Two, while do ... while loop

Program 3: 3N + 1 problem

The following procedures Bug

#include<stdio.h>
int main(){
    int n,count = 0;
    scanf("%d",&n);
    while(n >1){
        if(n%2 ==1)
            n = n *3 +1;
        else
            n /= 2;
        count++;
    }
    printf("%d",count);
    return 0;
}

When a large number of inputs, such as 987654321, the final output is a

A "" outputs an intermediate result "error method

#include<stdio.h>
int main(){
    int n,count = 0;
    scanf("%d",&n);
    while(n >1){
        if(n%2 ==1){
            n = n *3 +1;
            printf(" %d\n",n);
        }else{
            n /= 2;
        }
        count++;
    }
    printf("%d",count);
    return 0;
}

Found ----> multiplication overflow

# C99 does not specify the exact size of type int, but in the popular contest platform, int is 32 ---- -2147483648 to 2147483647

And the upper bound of n in question present upper limit of 10 ^ 9 is slightly smaller than the int, very easy to overflow

Therefore, problems can be solved using the long long - are: -2 ^ 63 to 2 ^ 63 - 1% lld input

Follow discussed in detail!

#include<stdio.h>
int main(){
    int n2,count = 0;
    scanf("%d",&n2);
    long long n = n2;
    while(n >1){
        if(n%2 ==1)
            n = n *3 +1;
        else
            n /= 2;
        count++;
    }
    printf("%d",count);
    return 0;
}

Procedure 4: approximated

#include<stdio.h>
int main(){
    double sum = 0;
    for(int i = 0;;i++){
        double term = 1.0 / (i*2 + 1);
        if(i%2 ==0)
            sum += term;
        else
            sum -= term;
        if(term < 1e-6)
            break;
    }
    printf("%.6f\n",sum);
    return 0;
}

#include<stdio.h>
int main(){
    double sum = 0;
    int i = 0;
    double term = 0;
    do{
        term = 1.0/(i*2+1);
        if(i%2 == 0)
            sum += term;
        else
            sum -= term;
        i++;
    }while(term > 1e-6);
    printf("%.6f",sum);
    return 0;
}

Third, the cost of recycling

Program 5: factorial sum

#include <stdio.h>
 int main () {
     int n-, S = 0 ; 
    Scanf ( " % D " , & n-);
     for ( int I = . 1 ; I <= n-; I ++ ) {
         int factorial = . 1 ; // at the beginning of the loop defined variables, each time execution of the loop are declared and initialized patronize 
        for ( int J = . 1 ; J <= I; J ++ ) 
            factorial * = J; 
        S + = factorial; 
    } 
    the printf ( " % D \ n- " , S%1,000,000 ); // for as long as the end 6, it is necessary for the modulo 10 ^ 6 output 
    return  0 ; 
}

Obviously, very easy to overflow

When n = 10 ^ 6, the overflow will be more, but very slow

NOTE: To calculate contains only integer addition, subtraction and multiplication expression is divided by a positive integer n, n may take the remainder after each step in the calculation, the same results

Next, the program into the "modulo every step" form, and then add a timer to see Speed:

#include<stdio.h>
#include<time.h>
int main(){
    const int MOD = 1000000;
    int n,s = 0;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
        int factorial = 1;
        for(int j = 1;j <=i;j++){
            factorial = (factorial * j %MOD);

        }
        s = (s + factorial) %MOD;
    }
    printf("%d\n",s);
    printf("Time used = %.2f\n",(double) clock() /CLOCKS_PER_SEC);
    return 0;
}

Timing function Clock () ---- so far run time of the function returns the program, call this function before the end of the program, you can get the entire running time of the program, after then divided by a constant CLOCKS_PER_SEC, the resulting value in seconds for the s unit

Keyboard input time is also taken into account, in order to avoid data input time affect the test results, you can use a technique called "pipe" tips:

Perform echo 20 in windows command line | abc, the system will automatically input 20, abc is the name of the program

Fourth, the input to the algorithm in the race, the output of the frame

Program 6: statistics

Some input integer, min max obtained and their average (three decimal places). Enter the number to ensure that these are no more than 1,000 integer

[There are] Bug

#include<stdio.h>
int main(){
    int x,n =0,min,max,s = 0;
    while(scanf("%d",&x) == 1){
        s += x;
        if(x <min)
            min = x;
        if(x >max)
            max = x;
        n++;
    }
    printf("%d %d %.3f",min,max,(double) s / n);
    return 0;
}

# Scanf () returns the number of input variables of success

have a test

5,002,320 come from? ---- variable value no previous assignment is uncertain !

The solution is to use before initial value

A better way is to use the file ---- input data stored in the file, the output data is also saved in the file. As long as the previously entered data stored in the file, there is no need to re-enter the

The easiest way to use the file is to use the input and output redirection

Only need to add the entrance to the main function:

freopen("input.txt","r",stdin)
freopen("output.txt","w",stdout)

At such a statement scanf () reads the write output.txt from input.txt, printf ()

Algorithm contest, players should strictly abide by the provisions of the file name of the game, especially [path can not add the path, even if it is a relative path]

Methods: using file redirection when the unit test, but once the game is automatically submitted to "" delete "" Redirect statement

#define the LOCAL 
#include <stdio.h>
 #define INF 1000000000
 int main () { 
    #ifdef the LOCAL 
        The freopen ( " data.in. " , " R & lt " , stdin); 
        The freopen ( " data.out " , " W " , stdout ); 

    #endif  // the LOCAL 
    int X, = n- 0 , min = INF, max = -INF, S = 0 ; // Causes of INF is: given a virtual infinity of
     the while (Scanf ( " % D " , X &) == . 1 ) {
        s += x;
        if(x <min)
            min = x;
        if(x >max)
            max= x;
        /*
        printf("x = %d,min = %d,max = %d\n",x,min,max);
        */
        n++;
    }
    printf("%d %d %.3f\n",min,max,(double) s/ n);

    return 0;
}

If the game requires a file input and output, but the way prohibit the use of redirection:

#include<stdio.h>
#define INF 1000000000
int main(){
    FILE *fin,*fout;
    fin = fopen("data.in","r");
    fout = fopen("data.out","wb");
    int x,n = 0,min = INF,max = -INF,s = 0;
    while(fscanf(fin,"%d",&x)== 1){
        s += x;
        if(x < min)
            min = x;
        if(x >max)
            max = x;
        n++;
    }
    fprintf(fout,"%d %d %.3f\n",min,max,(double) s / n);
    fclose(fin);
    fclose(fout);
    return 0;
}

用fopen("con","r")的方法打开标准输入输出不是可移植的，在Linux下是无效的！！！！

程序7：多组数据问题

#include<stdio.h>
#define INF 1000000000
int main(){
    int x,n =0,min = INF,max = -INF,s = 0,kase = 0;
    while(scanf("%d",&n) == 1&& n){
        int s= 0;
        for(int i = 0;i <n;i++){
            scanf("%d",&x);
            s += x;
            if(x < min)
                min = x;
            if(x >max)
                max = x;

        }
        if(kase)
            printf("\n");
        printf("Case %d: %d %d %.3f\n",++kase,min,max,(double) s/n);

    }
    return 0;
}

要点分析：