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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
spoondevil
Sample Output
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;
}