UVA12086 - Potentiometer(线段树/树状数组)

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
;
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;
}