C++——USACO Section 1.5 题解
Number Triangles
Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.
PROGRAM NAME: numtri
INPUT FORMAT
The first line contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.SAMPLE INPUT (file numtri.in)
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
OUTPUT FORMAT
A single line containing the largest sum using the traversal specified.SAMPLE OUTPUT (file numtri.out)
#include
int f[1001][1001],ans;
inline int readint()
{
int i=0;
char ch;
for(ch=getchar();ch<'0'||ch>'9';ch=getchar());
for(;ch>='0' && ch<='9';ch=getchar())
i=(i<<3)+(i<<1)+ch-'0';
return i;
}
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n;
n=readint();
for(int i=1;i<=n;++i)
for(int j=1;j<=i;++j)
f[i][j]=readint();
for(int i=1;i<=n;++i)
for(int j=1;j<=i;++j)
f[i][j]+=max(f[i-1][j-1],f[i-1][j]);
for(int i=1;i<=n;++i) ans=max(ans,f[n][i]);
int num=0;
char c[12];
do
{
c[++num]=(ans%10)+48;
ans/=10;
}while(ans);
while(num) putchar(c[num--]);
return 0;
} Prime Palindromes
The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 100,000,000); both a and b are considered to be within the range .
PROGRAM NAME: pprime
INPUT FORMAT
| Line 1: | Two integers, a and b |
SAMPLE INPUT (file pprime.in)
5 500
OUTPUT FORMAT
The list of palindromic primes in numerical order, one per line.SAMPLE OUTPUT (file pprime.out)
5 7 11 101 131 151 181 191 313 353 373 383
#include
#include
#include
#include
FILE *fout;
long a, b;
int
isprime(long n)
{
long i;
if(n == 2)
return 1;
if(n%2 == 0)
return 0;
for(i=3; i*i <= n; i+=2)
if(n%i == 0)
return 0;
return 1;
}
void
gen(int i, int isodd)
{
char buf[30];
char *p, *q;
long n;
sprintf(buf, "%d", i);
p = buf+strlen(buf);
q = p - isodd;
while(q > buf)
*p++ = *--q;
*p = '\0';
n = atol(buf);
if(a <= n && n <= b && isprime(n))
fprintf(fout, "%ld\n", n);
}
void
genoddeven(int lo, int hi)
{
int i;
for(i=lo; i<=hi; i++)
gen(i, 1);
for(i=lo; i<=hi; i++)
gen(i, 0);
}
void
generate(void)
{
genoddeven(1, 9);
genoddeven(10, 99);
genoddeven(100, 999);
genoddeven(1000, 9999);
}
void
main(void)
{
FILE *fin;
fin = fopen("pprime.in", "r");
fout = fopen("pprime.out", "w");
assert(fin != NULL && fout != NULL);
fscanf(fin, "%ld %ld", &a, &b);
generate();
exit (0);
} Superprime Rib
Butchering Farmer John's cows always yields the best prime rib. You can tell prime ribs by looking at the digits lovingly stamped across them, one by one, by FJ and the USDA. Farmer John ensures that a purchaser of his prime ribs gets really prime ribs because when sliced from the right, the numbers on the ribs continue to stay prime right down to the last rib, e.g.:
7 3 3 1
The set of ribs denoted by 7331 is prime; the three ribs 733 are prime; the two ribs 73 are prime, and, of course, the last rib, 7, is prime. The number 7331 is called a superprime of length 4.
Write a program that accepts a number N 1 <=N<=8 of ribs and prints all the superprimes of that length.
The number 1 (by itself) is not a prime number.
PROGRAM NAME: sprime
INPUT FORMAT
A single line with the number N.SAMPLE INPUT (file sprime.in)
4
OUTPUT FORMAT
The superprime ribs of length N, printed in ascending order one per line.SAMPLE OUTPUT (file sprime.out)
2333 2339 2393 2399 2939 3119 3137 3733 3739 3793 3797 5939 7193 7331 7333 7393
#include
#include
#include
#include
FILE *fout;
int
isprime(int n)
{
int i;
if(n == 2)
return 1;
if(n%2 == 0)
return 0;
for(i=3; i*i <= n; i+=2)
if(n%i == 0)
return 0;
return 1;
}
/* print all sprimes possible by adding ndigit digits to the number n */
void
sprime(int n, int ndigit)
{
if(ndigit == 0) {
fprintf(fout, "%d\n", n);
return;
}
n *= 10;
if(isprime(n+1))
sprime(n+1, ndigit-1);
if(isprime(n+3))
sprime(n+3, ndigit-1);
if(isprime(n+7))
sprime(n+7, ndigit-1);
if(isprime(n+9))
sprime(n+9, ndigit-1);
}
void main(void)
{
int n;
FILE *fin;
fin = fopen("sprime.in", "r");
assert(fin != NULL);
fout = fopen("sprime.out", "w");
assert(fout != NULL);
fscanf(fin, "%d", &n);
sprime(2, n-1);
sprime(3, n-1);
sprime(5, n-1);
sprime(7, n-1);
exit (0);
}