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UVa 1644 Prime Gap (水题,暴力)
题意:给定一个数 n,求它后一个素数和前一个素数差。
析:先打表,再二分查找。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define print(a) printf("%d\n", (a))#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1300000 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int a[maxn];vector<int> prime;int main(){ memset(a, 0, sizeof(a)); m = (int)sqrt(maxn + 0.5); for(int i = 2; i <= m; i++) if(!a[i]) for(int j = i * i; j < maxn; j += i) a[j] = 1; for(int i = 2; i < maxn; ++i) if(!a[i]) prime.push_back(i); while(cin >> n && n){ if(!a[n]) printf("0\n"); else{ int pos = lower_bound(prime.begin(), prime.end(), n) - prime.begin(); printf("%d\n", prime[pos]-prime[pos-1]); } } return 0;}
UVa 1644 Prime Gap (水题,暴力)
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