玲珑学院OJ 1005 - Spoon Devil's RP Test（求余问题）

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题目链接：http://www.ifrog.cc/acm/problem/1005

1005 - Spoon Devil’s RP Test
Time Limit：1s Memory Limit：32MByte

Submissions：83Solved：49

DESCRIPTION
Spoon Devil finds a way to test one person’s RP: He defines ‘a’ = 1, ‘b’ = 2^2, … ‘z’ = 26^2, so the value of ‘abc’ is 149, and the RP of ‘abc’ is the value of ‘abc’ mod 101. So the RP of ‘abc’ is 48.

INPUT
The first line is a single integer
T
T which is the number of test cases.
Each case only contains a name, which only contains lower-case letter.
OUTPUT
For each test case, output is the RP of the name in one line.
SAMPLE INPUT
1
spoondevil
SAMPLE OUTPUT
78
SOLUTION
“玲珑杯”acm比赛-试运行赛

【题意】给你每个字母的RP值，让你算出总的RP值，然后对101求余。
【思路】每加一个字母求一次余，暴力扫一遍就OK了。

下面是AC代码：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;char a[105];int num(int n)//计算数字位数{    int sum=0;    while(n)    {        n/=10;        sum++;    }    return sum;}int power(int n)//计算10的次幂{    int ans=1;    for(int i=0;i<n;i++)    {        ans*=10;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%s",a);        int len=strlen(a);        int rp=0,l=0;        for(int i=0;i<len;i++)        {            rp=rp*power(num((a[i]-'a'+1)*(a[i]-'a'+1)))+(a[i]-'a'+1)*(a[i]-'a'+1);            rp%=101;        }        printf("%d\n",rp);    }    return 0;}

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