Fortun e Test

Fortun e Test

时间： 1ms        内存：32M

描述：

A lot of people have played fortune Test(or called RP Test) before. Let ’ s create a new Test below.

Suppose the worth of a=1, b=4,c=9 … and z=26^2. Then abc can describe as 149,and abd describe as

1416. As it is too large ,we take it mod 101 as ours fortune worth. S o abd has only 2 RP. Now I give you

a name, please tell me the worth of it.

输入：

The first line of the input contains the number of test cases in the file. Each test case that follows

consists of one lines. each case contains only one string s specifying a person ’ s name, which only

contains lower-case .

输出：

For each test case, print a line contains the answer.

示例输入：

1
a

示例输出：

1

提示：

参考答案（内存最优[752]）：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int mod10(int x)
{
	int a=1,i;
	for(i=0;i<x;i++)
		a=(a*10)%101;
	return a;
}

int bit(char c)
{
	if(c>'C')
	{
		if(c>'I')
			return 3;
		else
			return 2;
	}
	else
		return 1;
}

int main()
{
	int i,j,k,l,len,n,t,sum;
	int shu[1000];
	char s[100];
	scanf("%d",&t);
	while(t--)
	{
		n=0;sum=0;
		scanf("%s",s);
		len=strlen(s);
		for(i=0;i<len;i++)
		{
			if(s[i]>='a')
				s[i]-=32;
		}
		for(i=0,j=0;i<len;i++)
		{
			n=(s[i]-'A'+1)*(s[i]-'A'+1);
			k=bit(s[i]);
			j+=k;
			for(l=1;l<=k;l++)
			{
				shu[j-l]=n%10;
				n/=10;
			}
		}
		for(i=j-1,k=0;i>=0;i--,k++)
			sum+=(shu[i]*mod10(k))%101;
		printf("%d\n",sum%101);
	}
	return 0;
}

参考答案（时间最优[0]）：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int mod10(int x)
{
	int a=1,i;
	for(i=0;i<x;i++)
		a=(a*10)%101;
	return a;
}

int bit(char c)
{
	if(c>'C')
	{
		if(c>'I')
			return 3;
		else
			return 2;
	}
	else
		return 1;
}

int main()
{
	int i,j,k,l,len,n,t,sum;
	int shu[1000];
	char s[100];
	scanf("%d",&t);
	while(t--)
	{
		n=0;sum=0;
		scanf("%s",s);
		len=strlen(s);
		for(i=0;i<len;i++)
		{
			if(s[i]>='a')
				s[i]-=32;
		}
		for(i=0,j=0;i<len;i++)
		{
			n=(s[i]-'A'+1)*(s[i]-'A'+1);
			k=bit(s[i]);
			j+=k;
			for(l=1;l<=k;l++)
			{
				shu[j-l]=n%10;
				n/=10;
			}
		}
		for(i=j-1,k=0;i>=0;i--,k++)
			sum+=(shu[i]*mod10(k))%101;
		printf("%d\n",sum%101);
	}
	return 0;
}

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