HOJ 11107

Number of stones

Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:32768KB

Total submit users: 13, Accepted users: 1

Problem 11107 : No special judgement

Problem description

There are N baskets rounded in a circle, and numbered as 1、2、3、…….、N-1、N, in clockwise. At the beginning, all of the baskets are empty. Some workers go to the moutain to collect stones. When they are back,they put their stones to some baskets. The workers have a habit, Once a worker come back, he choose a baskets, and choose a direction(clockwise or anticlockwise), he put one stone to this basket and move to the next basket according to the direction he has chosen, he continues doing this until all of the stones they have collected have been put down. Sometimes the workers ask you how many stone it is in the basket, as there are too many baskets, You are to wirte a program to calculate this.

Input

The input test file will contain multiple cases. Each test case: In the first line of the input contain two integers N,M(3 <= N <= 100000, 0 <= M <= 300000). Following M lines,each line represents an event. There are only three kinds of events: Q, C, U. And the format is: “Q x”, query the number of stones in the basket x. “C x num”, a worker comes back and the number of the stones he has picked up is num, he puts down stones from the basket x in clockwise. “U x num”, a worker comes back and the number of the stone he has picked up is num, he puts down stones from the basket x in anticlockwise. (x, num are both integers, 1 <= x <= N, 1 <= num <= 10000)

Output

For each query “Q x”, print the current number of stones in basket x.

Sample Input

5 8 C 5 8 Q 5 Q 4 U 5 3 C 2 6 Q 2 Q 1 U 2 8

Sample Output

2 1 4 3

Problem Source

birdman

上次比赛没有做..补做一个..挺好的题..重写了点树模板

 1 /**/ /*
 2 * 主程序要作的事情
 3 * 1.确定N ：必须是2^n，可以取实际n的上界
 4 * 2.build(left, right);
 5 *
 6 */
 7
 8 #include  < cstdio >
 9 #include  < cstring >
10
11 const   int  N  =   131072 ;                 // 必须是2^n，可以取实际n的上界
12
13 int  upperbound;
14
15 struct Node  {
16    int i, j, c, m;                    //left, right
17}  T[N * 2 ];
18
19 void  bd( int  d,  int  left,  int  right)  {
20    T[d].i = left, T[d].j = right, T[d].c = 0;
21    if(right > left) {
22        bd(d*2+1, left, T[d].m = (left+right)>>1);
23        bd(d*2+2, T[d].m+1, right);
24    }
25}
26
27 void  build( int  left,  int  right)  {
28    upperbound = 1;
29    while(upperbound < right-left+1) upperbound <<= 1;
30    bd(0, left, left + upperbound-1);
31}
32
33 void  add( int  d,  int  left,  int  right,  int  c)  {
34    if(left <= T[d].i && right >= T[d].j) {
35        T[d].c += c;
36    }
37    else {
38        if(left <= T[d].m) add(d*2+1, left, right, c);
39        if(right > T[d].m) add(d*2+2, left, right, c);
40    }
41}
42
43 int  get( int  x)  { // 获得点的覆盖次数
44    int idx = upperbound+x-1, sum = 0;
45    do {
46        sum += T[idx].c;
47        idx = (idx-1)>>1;
48    } while(idx != 0);
49    return sum;
50}
51
52 int  n, m;
53
54 void  Add( int  x,  int  num)  {
55    int laps = (num-(n-x))/n;
56    if(laps > 0) {
57        add(0, 0, n-1, laps);
58    }
59    num -= laps*n;
60    if(n-x >= num) {
61        add(0, x, x+num-1, 1);
62    }
63    else {
64        add(0, x, n-1, 1);
65        add(0, 0, (x+num-1)%n, 1);
66    }
67}
68
69 int  main()  {
70    while(scanf("%d %d", &n, &m) != EOF) {
71        build(0, n-1);
72        while(m--) {
73            char cmd;
74            int x, num;
75            scanf(" %c", &cmd);
76            if(cmd == 'Q') {
77                scanf("%d", &x); 
78                --x;
79                printf("%d\n", get(x));
80            }
81            else if(cmd == 'C') {
82                scanf("%d %d", &x, &num);
83                --x;
84                Add(x, num);
85            }
86            else if(cmd == 'U') {
87                scanf("%d %d", &x, &num);
88                --x;
89                int y = (x-num+1) % n;
90                if(y < 0) y += n;
91                Add(y, num);
92            }
93        }
94    }
95
96    return 0;
97}