ZOJ 3785 What day is that day? 数论
Posted ( m Lweleth)
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LINK: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3785
题意:求 \(1^1+2^2+3^3+…+n^n\) 加1模7的值
思路:其实这题是找规律题...说数论是不甘心,这题卡了好久,借助python打表发现 指数每6循环 n值每7循环 6*7=42 但每42个数后它们的排列就会发生改变,共改变7次,所以总循环周期为42*7=294,预处理记录一下就好。
/** @Date : 2017-03-23-23.08 * @Author : Lweleth ([email protected]) * @Link : https://github.com/ * @Version : */ #include<bits/stdc++.h> #define LL long long #define PII pair #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-8; int rec[300]; LL sum[300]; char day[8][10] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" , "Saturday", "Sunday"}; int main() { rec[0] = 0; sum[0] = 0; for(int i = 1; i <= 42*7; i++) { int r; if(i % 7 == 0) r = 0; else r = (int)pow(i%7, i%6) % 7; rec[i] = r; sum[i] = sum[i - 1] + rec[i]; } int T; LL n; cin >> T; while(T--) { scanf("%lld", &n); LL x = n / (42*7); LL y = n % (42*7); LL ans = (x * sum[294] + sum[y] + 6) % 7; printf("%s\n", day[ans]); } return 0; }
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