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
", 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 ("
");
printf ("Case %d:
", ++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
", 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 ("
");
memset (C, 0, sizeof (C));
for (int i = 1; i <= n; i++) {
scanf ("%d", &A[i]);
Add(i, A[i]);
}
printf ("Case %d:
", ++cas);
solve();
}
return 0;
}