HDU 5047 Sawtooth(数学 公式 大数)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5047

Problem Description

Think about a plane:

● One straight line can divide a plane into two regions.
● Two lines can divide a plane into at most four regions.
● Three lines can divide a plane into at most seven regions.
● And so on...

Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?

 

Input

The first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.

Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 10 ¹²)

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then an integer that is the maximum number of regions N the “M” figures can divide.

 

Sample Input

2 1 2

 

Sample Output

Case #1: 2 Case #2: 19

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

PS:

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代码如下：

#include 
#include 
#include 
/*大数加法*/
void add(char* a,char* b,char* c)
{
    int i,j,k,max,min,n,temp;
    char *s,*pmax,*pmin;
    max=strlen(a);
    min=strlen(b);
    if (max=0; i--,j--,k--)
        s[k]=pmin[i]-'0'+pmax[j];
    for (; j>=0; j--,k--)
        s[k]=pmax[j];
    for (i=max; i>=0; i--)
        if (s[i]>'9')
        {
            s[i]-=10;
            s[i-1]++;
        }
    if (s[0]=='0')
    {
        for (i=0; i<=max; i++)
            c[i-1]=s[i];
        c[i-1]='\0';
    }
    else
    {
        for (i=0; i<=max; i++)
            c[i]=s[i];
        c[i]='\0';
    }
    free(s);
}

/*大数减法*/
void subtract(char* a,char* b,char* c)
{
    int i,j,ca,cb;
    ca=strlen(a);
    cb=strlen(b);
    if (ca>cb||(ca==cb&&strcmp(a,b)>=0))
    {
        for (i=ca-1,j=cb-1; j>=0; i--,j--)
            a[i]-=(b[j]-'0');
        for (i=ca-1; i>=0; i--)
            if (a[i]<'0')
            {
                a[i]+=10;
                a[i-1]--;
            }
        i=0;
        while (a[i]=='0')
            i++;
        if (a[i]=='\0')
        {
            c[0]='0';
            c[1]='\0';
        }
        else
        {
            for (j=0; a[i]!='\0'; i++,j++)
                c[j]=a[i];
            c[j]='\0';
        }
    }
    else
    {
        for (i=ca-1,j=cb-1; i>=0; i--,j--)
            b[j]-=(a[i]-'0');
        for (j=cb-1; j>=0; j--)
            if (b[j]<'0')
            {
                b[j]+=10;
                b[j-1]--;
            }
        j=0;
        while (b[j]=='0')
            j++;
        i=1;
        c[0]='-';
        for (; b[j]!='\0'; i++,j++)
            c[i]=b[j];
        c[i]='\0';
    }
}

/* 大数乘法*/
void multiply(char* a,char* b,char* c)
{
    int i,j,ca,cb,* s;
    ca=strlen(a);
    cb=strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));
    for (i=0; i=0; i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }
    i=0;
    while (s[i]==0)
        i++;
    for (j=0; i