UVA12086 - Potentiometer(线段树/树状数组)

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题目链接

题目大意：给你N个数字，然后有q个操作，操作类型：M代表修改某个位置的值为r，S代表查询某一段的数字和。

解题思路：线段树或者树状数组。

线段树

#include <cstdio>#include <cstring>const int N = 8e5 + 5;int v[N];int n;int Query (int o, int l, int r, int ql, int qr) {    int    m = l + (r - l) / 2;    if (ql == l && r == qr)        return v[o];     if (qr <= m)        return Query(2 * o, l, m, ql, qr);    else if (ql > m)        return Query(2 * o + 1, m + 1, r, ql, qr);    else         return Query(2 * o, l, m, ql, m) + Query(2 * o + 1, m + 1, r, m + 1, qr);}void Update (int o, int l, int r, int p, int val) {    int m = l + (r - l) / 2;    if (l == r)         v[o] = val;    else {        if (p <= m)            Update (2 * o, l, m, p, val);        else            Update (2 * o + 1, m + 1, r, p, val);        v[o] = v[2 * o] + v[2 * o + 1];    }}void solve () {    char str[10];    int x, y, r;    while (scanf ("%s", str) && str[0] != 'E') {        if (str[0] == 'M') {            scanf ("%d%d", &x, &y);            printf ("%d\n", Query (1, 1, n, x, y));        } else {            scanf ("%d%d", &x, &r);            Update (1, 1, n, x, r);            }    }}int main () {    int cas = 0;    int num;    while (scanf ("%d", &n) && n) {        if (cas)            printf ("\n");        printf ("Case %d:\n", ++cas);        memset (v, 0, sizeof (v));        for (int i = 1; i <= n; i++) {            scanf ("%d", &num);            Update (1, 1, n, i, num);        }        solve();                }    return 0;}

树状数组

#include <cstdio>#include <cstring>const int maxn = 2e5 + 5;int lowbit (int x) { return x&-x; }int n;int A[maxn], C[maxn];void Add (int x, int d) {    while (x <= n) {        C[x] += d;        x += lowbit(x);    }}int Sum (int x) {    int ret = 0;    while (x > 0) {        ret += C[x];        x -= lowbit(x);    }    return ret;}void solve () {    char str[10];    int x, r, y;    while (scanf ("%s", str) && str[0] != 'E') {        if (str[0] == 'M') {            scanf ("%d%d", &x, &y);            printf ("%d\n", Sum (y) - Sum (x - 1));        } else {            scanf ("%d%d", &x, &r);            Add(x, r - A[x]);            A[x] = r;//这个地方要记得修改        }    }}int main () {    int cas = 0;    while (scanf ("%d", &n) && n) {        if (cas)            printf ("\n");        memset (C, 0, sizeof (C));        for (int i = 1; i <= n; i++) {            scanf ("%d", &A[i]);            Add(i, A[i]);        }        printf ("Case %d:\n", ++cas);        solve();    }    return 0;}

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