数学 Codeforces Round #282 (Div. 2) B. Modular Equations
题意:a % x == b,求符合条件的x有几个
数学:等式转换为:a == nx + b,那么设k = nx = a - b,易得k的约数(>b)的都符合条件,比如a=25 b=1,那么24,12, 8, 6, 4, 3, 2都可以,所以只要求出k的约数有几个就可以了,a <= b的情况要特判
/************************************************
* Author :Running_Time
* Created Time :2015-8-19 18:49:21
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int divisor(int n, int b) {
int ret = 0;
for (int i=1; i*i<=n; ++i) {
if (n % i == 0) {
if (i > b) ret++;
if (n / i != i && n / i > b) ret++;
}
}
return ret;
}
int main(void) { //Codeforces Round #282 (Div. 2) B. Modular Equations
int a, b; scanf ("%d%d", &a, &b);
if (a < b) {
puts ("0"); return 0;
}
if (a == b) {
puts ("infinity"); return 0;
}
int ans = divisor (a - b, b);
printf ("%d\n", ans);
return 0;
}
* Author :Running_Time
* Created Time :2015-8-19 18:49:21
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int divisor(int n, int b) {
int ret = 0;
for (int i=1; i*i<=n; ++i) {
if (n % i == 0) {
if (i > b) ret++;
if (n / i != i && n / i > b) ret++;
}
}
return ret;
}
int main(void) { //Codeforces Round #282 (Div. 2) B. Modular Equations
int a, b; scanf ("%d%d", &a, &b);
if (a < b) {
puts ("0"); return 0;
}
if (a == b) {
puts ("infinity"); return 0;
}
int ans = divisor (a - b, b);
printf ("%d\n", ans);
return 0;
}
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