暑期多校联合比赛F题 8_24 杨辉三角的打印

F、Pascal'sTriangle of DeathTime                 limit: 1.000 seconds

In this problem, you are asked to generate Pascal's Triangle.

Pascal's Triangle is useful in many areas from probability to

polynomials to programming contests. It is a triangle of integers

with ``1'' on top and down the sides. Any number in the interior

equals the sum of the two numbers above it. For example, here are

the first 5 rows of the triangle.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

In ``Pascal's Triangle of Death,'' you are to generate a left

justified Pascal's Triangle. When any number in the triangle is

exceeds or equals 10^60, your program should finish printing the

current row and exit. The output should have each row of the

triangle on a separate line with one space between each element.

The final element of each line should be directly followed by a

newline. There is no space after the last number on each line.

Sample Input

There is no input for this problem.

Sample Output

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

.

.

.

etc.

题目大意：打印杨辉三角直到遇到一个等于或者大于10^60的数(当然还有接着打印完这一行)，一开始被这大数吓倒了，尼玛可恶啊！

思路：其实我就卡在数学的杨辉三角没学好，没算到大约205层就可以到10^60这个数了，

还有并不知道三维数组竟然可以开到那么大，我勒个去啊，对不起队友羽哥还有杨烁啊 。

 

 

 

program：

 

#include <stdio.h> 

#include <string.h> 

 

#define MAX 300 

char a[MAX][MAX][MAX]; 

void reverse(char *from, char *to); 

void call_sum(char *first, char *second, char *result); 

int main() 

{ 

 char res[MAX]; 

 int flag = 1; 

 int count, i, len; 

 strcpy(a[0][0], "1"); 

 strcpy(a[1][0], "1"); 

 strcpy(a[1][1], "1"); 

 printf("1\n1 1\n"); 

 count = 2; 

while (flag)

{ 

  strcpy(a[count][0], "1"); 

  strcpy(a[count][count], "1"); 

   for (i = 1; i < count; i++)

   { 

     call_sum(a[count - 1][i - 1], a[count - 1][i], res); 

     len = strlen(res); 

     strcpy(a[count][i], res); 

     if (len >= 61) 

       flag = 0; 

   } 

   for (i = 0; i <= count; i++)

    { 

        printf("%s", a[count][i]); 

        if (i < count) 

        printf(" ");  //打印该行

        else 

        printf("\n"); 

    } 

  count++; 

} 

 

 return 0; 

} 

 

void reverse(char *from, char *to) 

{ 

 int len = strlen(from); 

 int i; 

 for (i = 0; i < len; i++) { 

  to[i] = from[len - 1 - i]; 

 to[len] = '\0'; 

 } 

} 

void call_sum(char *first, char *second, char *result) 

{ 

 char F[MAX], S[MAX], Res[MAX]; 

 int f, s, sum, extra, now; 

 f = strlen(first); 

 s = strlen(second); 

 reverse(first, F); 

 reverse(second, S); 

     for (now = 0, extra = 0; now < f && now < s; now++)

      { 

        sum = (F[now] - '0') + (S[now] - '0') + extra; 

        Res[now] = sum % 10 + '0'; 

        extra = sum / 10; 

       } 

     for (; now  < f; now++)

     { 

        sum = F[now] + extra - '0'; 

        Res[now] = sum % 10 + '0'; 

        extra = sum / 10; 

     } 

     for (; now < s; now++)

      { 

        sum = S[now] + extra -'0'; 

        Res[now] = sum % 10 + '0'; 

        extra = sum / 10; 

      } 

 if (extra) 

    Res[now++] = extra + '0';  //最高位多算一次

 Res[now] = '\0';//去掉前导0 

 if (strlen(Res) == 0) 

    strcpy(Res, "0"); 

 reverse(Res, result); 

}