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

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
#include
#include
#include
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;i10;
    }
    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'a'+1)*(a[i]-'a'+1)))+(a[i]-'a'+1)*(a[i]-'a'+1);
            rp%=101;
        }
        printf("%d\n",rp);
    }
    return 0;
}