题目链接：
http://codeforces.com/contest/675/problem/A
题意：
给你一个等差数列，问你b是不是其中的一项。
即是否存在大于等于0的整数k使得a+k*c==b。只要求(b-a)%c==0&&(b-a)/c>=0就可以了。
但是，我就是写成了二分。。而且还写搓了。。
先看看哇的代码：
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 7 const int maxn = 333; 8 const int INF = 1e9; 9 10 int main() { 11 int a, b, c; 12 while (scanf("%d%d%d", &a, &b, &c) == 3) { 13 if (c == 0) { 14 if (b == a) puts("YES"); 15 else puts("NO"); 16 continue; 17 } 18 int low, hig,flag; 19 if (c < 0) { 20 flag = 0; 21 low=0, hig = (-INF - a) / c + 10; 22 if (b > a) { 23 puts("NO"); continue; 24 } 25 while (low < hig-1) { 26 int mid = low + (hig - low) / 2; 27 if (a + mid*c>b) { 28 low = mid + 1; 29 } 30 else if (a + mid*c < b) { 31 hig = mid; 32 } 33 else { 34 flag = 1; break; 35 } 36 } 37 if (flag) puts("YES"); 38 else puts("NO"); 39 } 40 else if (c>0) { 41 flag = 0; 42 low = 0, hig = (INF - a) / c + 10; 43 if (b < a) { 44 puts("NO"); continue; 45 } 46 while (low < hig - 1) { 47 int mid = low + (hig - low) / 2; 48 if (a + mid*c<b) { 49 low = mid + 1; 50 } 51 else if (a + mid*c > b) { 52 hig = mid; 53 } 54 else { 55 flag = 1; break; 56 } 57 } 58 if (flag) puts("YES"); 59 else puts("NO"); 60 } 61 } 62 return 0; 63 }
以后写二分以此为鉴。。
贴修改之后的：
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 7 const int maxn = 333; 8 typedef long long LL; 9 const LL INF = 1e9; 10 11 int main() { 12 int a, b, c; 13 while (scanf("%d%d%d", &a, &b, &c) == 3) { 14 if (c == 0) { 15 if (b == a) puts("YES"); 16 else puts("NO"); 17 continue; 18 } 19 LL low, hig; 20 if (c < 0) { 21 low = 0, hig = (-INF - a) / c + 10; 22 if (b > a) { 23 puts("NO"); continue; 24 } 25 while (low < hig - 1) { 26 LL mid = low + (hig - low) / 2; 27 if (a + mid*c>=b) { 28 low = mid; 29 } 30 else if (a + mid*c < b) { 31 hig = mid; 32 } 33 } 34 if (a + low*c == b) puts("YES"); 35 else puts("NO"); 36 } 37 else if (c>0) { 38 low = 0, hig = (INF - a) / c + 10; 39 if (b < a) { 40 puts("NO"); continue; 41 } 42 while (low < hig - 1) { 43 LL mid = low + (hig - low) / 2; 44 if (a + mid*c<=b) { 45 low = mid; 46 } 47 else if (a + mid*c > b) { 48 hig = mid; 49 } 50 } 51 if (a + low*c == b) puts("YES"); 52 else puts("NO"); 53 } 54 } 55 return 0; 56 }