UVA12086 - Potentiometer(线段树/树状数组)

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;
}