线段树(单点更新) POJ 2886 Who Gets the Most Candies?

 

题目传送门

 1 #include <cstdio>  2 #include <cstring>  3 #define lson l, m, rt << 1  4 #define rson m+1, r, rt << 1 | 1  5  6 const int MAX_N = 500000 + 10;  7 int sum[MAX_N << 2];  8 struct node  9 { 10 char name[15]; 11 int val; 12 }boy[MAX_N<<2]; 13 int ans[MAX_N]; 14 int id; 15 int n, k; 16 17 void build(int l, int r, int rt) 18 { 19 sum[rt] = r - l + 1; 20 if (l == r) 21 return ; 22 int m = (l + r) >> 1; 23  build (lson); 24  build (rson); 25 } 26 27 int update(int p, int l, int r, int rt) 28 { 29 sum[rt]--; 30 if (l == r) 31  { 32 return l; 33  } 34 int m = (l + r) >> 1; 35 if (p <= sum[rt<<1]) 36  { 37 return update (p, lson); 38  } 39 else 40  { 41 return update (p - sum[rt<<1], rson); 42  } 43 } 44 45 void Solve(){ //计算ans 46 memset(ans,0,sizeof(ans)); 47 for(int i=1;i<=n;i++){ 48 ans[i]++; 49 for(int j=2*i;j<=n;j+=i) 50 ans[j]++; 51  } 52 int max=ans[1]; 53 id=1; 54 for(int i=2;i<=n;i++) //找出第几个人跳出获得的糖最多 55 if(ans[i]>max){ 56 max=ans[i]; 57 id=i; 58  } 59 } 60 61 int main(void) //POJ 2886 Who Gets the Most Candies? 62 { 63 //freopen ("inF.txt", "r", stdin); 64 65 while (~scanf ("%d%d", &n, &k)) 66  { 67 build (1, n, 1); 68 for (int i=1; i<=n; ++i) 69  { 70 scanf ("%s%d", &boy[i].name, &boy[i].val); 71 //printf ("%s%d\n", boy[i].name, boy[i].val); 72  } 73  Solve(); 74 //int maxn = f(n); 75 int mod = sum[1]; 76 boy[0].val = 0; 77 int pos = 0; 78 //printf ("%d\n", id); 79 int m=id; 80 while (m--) 81  { 82 if (boy[pos].val > 0) 83 k=((k-1+boy[pos].val-1)%mod+mod)%mod+1; 84 else 85 k=((k-1+boy[pos].val)%mod+mod)%mod+1; 86 pos = update (k, 1, n, 1); 87 //printf ("%d ", pos); 88 mod = sum[1]; 89  } 90 printf ("%s %d\n", boy[pos].name, ans[id]); 91  } 92 }