Lonlife-ACM 1005 - Spoon Devil's RP Test(同余定理)——“玲珑杯”acm比赛-试运行赛

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Lonlife-ACM 1005 - Spoon Devil's RP Test

Accept: 0 Submit: 0
Time Limit: 1s Memory Limit : 32MByte

Problem 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 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

Hint

Problem Idea

解题思路:

【题意】
定义'a' = 1, 'b' = 2^2, ... 'z' = 26^2

所以'abc'=149

'abc'的RP为149%101=48

现在给你一个由小写字母构成的人名

问该人名的RP值为多少

【类型】
同余定理
【分析】

由同余定理可得

①(a%m+b%m)%m≡(a+b)%m

②((a%m)*(b%m))%m≡(a*b)%m

故一个数,如567,当对4取模时,可以转化为(((5%4)*10+6)%4*10+7)%4

同理,此题就是利用这种转化,只是因为字符值不一定是一位数,故不仅仅会*10,可能还会*100,甚至*1000

【时间复杂度&&优化】
O(strlen(s))

题目链接→Lonlife-ACM 1005 - Spoon Devil's RP Test

Source Code

/*Sherlock and Watson and Adler*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<bitset>#include<cmath>#include<complex>#include<string>#include<algorithm>#include<iostream>#define eps 1e-9#define LL long long#define PI acos(-1.0)#define bitnum(a) __builtin_popcount(a)using namespace std;const int N = 100005;const int M = 100005;const int inf = 1000000007;const int mod = 101;char s[N];int fun(int x){    return x<10?10:x<100?100:1000;}int main(){    int t,k,i,ans;    scanf("%d",&t);    while(t--)    {        scanf("%s",s);        for(ans=i=0;s[i]!='\0';i++)        {            k=s[i]-'a'+1;            k*=k;            ans=(ans*fun(k)+k)%mod;        }        printf("%d\n",ans);    }    return 0;}

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