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HDU 1037 Keep on Truckin’[解题报告] C++

Keep on Truckin’

问题描述 :

Boudreaux and Thibodeaux are on the road again . . .

"Boudreaux, we have to get this shipment of mudbugs to Baton Rouge by tonight!"

"Don’t worry, Thibodeaux, I already checked ahead. There are three underpasses and our 18-wheeler will fit through all of them, so just keep that motor running!"

"We’re not going to make it, I say!"

So, which is it: will there be a very messy accident on Interstate 10, or is Thibodeaux just letting the sound of his own wheels drive him crazy?


Input to this problem will consist of a single data set. The data set will be formatted according to the following description.

The data set will consist of a single line containing 3 numbers, separated by single spaces. Each number represents the height of a single underpass in inches. Each number will be between 0 and 300 inclusive.


There will be exactly one line of output. This line will be:


if the height of the 18-wheeler is less than the height of each of the underpasses, or:


otherwise, where X is the height of the first underpass in the data set that the 18-wheeler is unable to go under (which means its height is less than or equal to the height of the 18-wheeler).
The height of the 18-wheeler is 168 inches.


180 160 170



2011-12-20 16:13:50




# include <stdio.h>

int main ()
    int n ;
    while (~scanf ("%d", &n))
        if (n < 168) break ;

    if (n < 168) printf ("CRASH %d\n", n) ;
    else printf ("NO CRASH\n") ;
    return 0 ;

  1. “可以发现,树将是满二叉树,”这句话不对吧,构造的树应该是“完全二叉树”,而非“满二叉树”。

  2. #include <stdio.h>
    int main()
    int n,p,t[100]={1};
    for(int i=1;i<100;i++)
    t =i;
    printf("Printing order for 1 pages:nSheet 1, front: Blank, 1n");
    else {
    if(n%4) p=n/4+1;
    else p=n/4;
    int q=4*p;
    printf("Printing order for %d pages:n",n);
    for(int i=0;i<p;i++){
    printf("Sheet %d, front: ",i+1);
    if(q>n) {printf("Blank, %dn",t[2*i+1]);}
    else {printf("%d, %dn",q,t[2*i+1]);}
    printf("Sheet %d, back : ",i+1);
    if(q>n) {printf("%d, Blankn",t[2*i+2]);}
    else {printf("%d, %dn",t[2*i+2],q);}
    return 0;

  3. 问题3是不是应该为1/4 .因为截取的三段,无论是否能组成三角形, x, y-x ,1-y,都应大于0,所以 x<y,基础应该是一个大三角形。小三角是大三角的 1/4.

  4. for(int i=1; i<=m; i++){
    for(int j=1; j<=n; j++){
    dp = dp [j-1] + 1;
    if(s1.charAt(i-1) == s3.charAt(i+j-1))
    dp = dp[i-1] + 1;
    if(s2.charAt(j-1) == s3.charAt(i+j-1))
    dp = Math.max(dp [j - 1] + 1, dp );
    这里的代码似乎有点问题? dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false