CodeForces - 659A Round House （形成环）水

CodeForces - 659A Round House Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1are adjacent. Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b = 0, then Vasya prefers to walk beside his entrance. Illustration for n = 6, a = 2, b =  - 5. Help Vasya to determine the number of the entrance, near which he will be at the end of his walk. Input The single line of the input contains three space-separated integers n, a and b (1 ≤ n ≤ 100, 1 ≤ a ≤ n,  - 100 ≤ b ≤ 100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively. Output Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk. Sample Input Input 6 2 -5 Output 3 Input 5 1 3 Output 4 Input 3 2 7 Output 3 //题意：输入n,a,b; 有n个房子，现在小明在第a个房子出，如果b<0,那么小明逆时针走过|b|个房子，若b>0则顺时针走b个房子，问小明最终在第几个房子处。 //在赋值的时候直接顺序赋值，在输出时就好输出了。一个小技巧 #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<iostream> #define INF 0x3f3f3f3f #define ll long long #define IN __int64 #define N 110 #define M 1000000007 using namespace std; int s[N]; int main() { int n,a,b; int i,j; while(scanf("%d%d%d",&n,&a,&b)!=EOF) { if(b>0) { for(i=0;i<n;i++) { if(a>n) a=1; s[i]=a++; } } else { for(i=0;i<n;i++) { if(a) s[i]=a; else { a=n; s[i]=a; } a--; } } int p=abs(b)%n; printf("%d\n",s[p]); } return 0; }