TCO14 2B L3: AlwaysDefined，math，从余数入手

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题目：http://community.topcoder.com/stat?c=problem_statement&pm=13060&rd=15996

参考：http://apps.topcoder.com/wiki/display/tc/TCO+2014+Round+2B

将数按对W的余数进行分类。做题思维要非常清晰，难度不小，看题解看了半天才看懂。

代码：

#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <iostream>#include <sstream>#include <iomanip>#include <bitset>#include <string>#include <vector>#include <stack>#include <deque>#include <queue>#include <set>#include <map>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <cstring>#include <ctime>#include <climits>using namespace std;#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)typedef pair<int, int> pii;typedef long long llong;typedef pair<llong, llong> pll;#define mkp make_pair/*************** Program Begin **********************/vector<vector<int>> g;class AlwaysDefined {public:inline long long items(long long rem, long long upper, int W){// rem + (cnt - 1) * W < upperlong long numbers = upper - rem;return (numbers + W - 1) / W;}long long solve(int W, int rem, long long cnt){if (cnt <= 0) {return 0;}if (1 == rem) {return cnt;}long long ans = 0;for (int i = 0; i < g[rem].size(); i++) {int x = g[rem][i];long long smallest = rem + x * W;long long newrem = smallest / rem;ans += solve(W, newrem, items(x, cnt, rem));// rem 为一个周期}return ans;}long long countIntegers(long long L, long long R, int W) {long long res = 0;g.resize(W);for (int i = 0; i < g.size(); i++) {g[i].clear();}for (int rem = 1; rem < W; rem++) {for (int x = 0; x < rem; x++) {if ( (x * W) % rem == 0 ) {g[rem].push_back(x);}}}for (int rem = 1; rem < W; rem++) {res += solve(W, rem, items(rem, R + 1, W));res -= solve(W, rem, items(rem, L, W));}return res;}};/************** Program End ************************/

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