标签：return 进阶 树状 int res scanf d% 0x00 pos
清点人数
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 5e5 + 10;
int n, k, c[N]; //c为原序列的树状数组
int lowbit(int x) { return x & -x; }
//将原序列下标为pos的元素值增加x，改变相应的c数组的值
void update(int pos, int x) {
for (int i = pos; i <= n; i += lowbit(i)) c[i] += x;
}
//对原序列前pos项求和
int sum(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c[i];
return res;
}
int main() {
scanf("%d%d", &n, &k);
while (k --) {
char op[2]; //注意字符变量的输入方式
int m, p;
scanf("%s", &op);
if (op[0] == 'A') scanf("%d", &m), printf("%d\n", sum(m));
if (op[0] == 'B') scanf("%d%d", &m, &p), update(m, p);
if (op[0] == 'C') scanf("%d%d", &m, &p), update(m, -p);
}
return 0;
}
数列操作
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1e5 + 10;
int n, m, x, c[N]; //c为原序列的树状数组
int lowbit(int x) { return x & -x; }
//将原序列下标为pos的元素值增加x，改变相应的c数组的值
void update(int pos, int x) {
for (int i = pos; i <= n; i += lowbit(i)) c[i] += x;
}
//对原序列前pos项求和
int sum(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c[i];
return res;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++) scanf("%d", &x), update(i, x);
while (m --) {
int k, a, b;
scanf("%d%d%d", &k, &a, &b);
//sum(b) - sum(a - 1)即为区间[a, b]的和
if (k == 0) printf("%d\n", sum(b) - sum(a - 1));
else update(a, b);
}
return 0;
}
简单题
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1e5 + 10;
int n, m, c[N];
int lowbit(int x) { return x & -x; }
void update(int pos, int x) {
for (int i = pos; i <= n; i += lowbit(i)) c[i] += x;
}
int sum(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c[i];
return res;
}
int main() {
scanf("%d%d", &n, &m);
while (m --) {
int t, l, r;
scanf("%d", &t);
if (t == 1) scanf("%d%d", &l, &r), update(l, 1), update(r + 1, -1);
else scanf("%d", &l), printf("%d\n", sum(l) % 2);
}
return 0;
}
数星星Stars
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 2e4 + 10, MAXX = 32000 + 10;
int n, m, c[MAXX], h[N];
int lowbit(int x) { return x & -x; }
void update(int pos, int x) {
for (int i = pos; i <= MAXX; i += lowbit(i)) c[i] += x;
}
int sum(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c[i];
return res;
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i ++) {
int x, y;
scanf("%d%d", &x, &y);
x ++; //根据题意，x有可能为0
int t = sum(x);
h[t] ++;
update(x, 1);
}
for (int i = 0; i < n; i ++) printf("%d\n", h[i]);
return 0;
}
校门外的树
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 5e4 + 10;
int n, m, c1[N], c2[N];
int lowbit(int x) { return x & -x; }
void update1(int pos, int x) {
for (int i = pos; i <= n; i += lowbit(i)) c1[i] += x;
}
void update2(int pos, int x) {
for (int i = pos; i <= n; i += lowbit(i)) c2[i] += x;
}
int sum1(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c1[i];
return res;
}
int sum2(int pos) {
int res = 0;
for (int i = pos; i > 0; i -= lowbit(i)) res += c2[i];
return res;
}
int main() {
scanf("%d%d", &n, &m);
while (m --) {
int k, l, r;
scanf("%d%d%d", &k, &l, &r);
if (k == 1) update1(l, 1), update2(r, 1);
else printf("%d\n", sum1(r) - sum2(l - 1));
}
return 0;
}
标签：return,进阶,树状,int,res,scanf,d%,0x00,pos 来源： https://blog.csdn.net/davidliule/article/details/122375988
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