标签：
zoj3741
简单dp。wa了两个小时，中间改了好多细节。后来还是不对，参考了别人的代码，发现一个致命问题，初始化的时候，不是每种状态都能直接达到的。初始化成-1。
（题目有个小坑，0<=L<=5, 即使吃药了，也不能到6 ）
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#define pk printf("lalala");
using namespace std;
#define PI acos(-1.0)
#define EXP exp(1.0) // 自然对数
#define ESP 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
typedef long long ll;
const int N = 105;
int dp[N][2*N];
int a[N];
void print()
{
for (int k = 1; k <= 6; ++k) {
for (int i = 98; i <= 101; ++i)
printf("%d ", dp[k][i]);
printf("\n");
}
}
int main()
{
int l, n, x, y;
while (~scanf("%d%d%d%d", &l, &n, &x, &y)) {
clr(dp, -1);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
dp[0][100] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 100 - y; j <= 100 + x; ++j) {
if (j == 100 - y) {
if (dp[i - 1][100 + x] == -1) continue;
if (0 >= a[i]) dp[i][j] = dp[i - 1][100 + x] + 1;
else dp[i][j] = dp[i - 1][100 + x];
} else if (j < 100) {
if (dp[i - 1][j - 1] == -1) continue;
if (0 >= a[i]) dp[i][j] = dp[i - 1][j - 1] + 1;
else dp[i][j] = dp[i - 1][j - 1];
} else if (j == 100 + 1) {
if (dp[i - 1][j - 1] == -1 && dp[i - 1][100 - 1] == -1) continue;
if ((l + 1 <= 5 && l + 1 >= a[i]) || l >= a[i]) {
dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][100 - 1]) + 1;
} else {
dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][100 - 1]);
}
} else if (j > 100) {
if (dp[i - 1][j - 1] == -1) continue;
if ((l + 1 <= 5 && l + 1 >= a[i]) || l >= a[i]) dp[i][j] = dp[i - 1][j - 1] + 1;
else dp[i][j] = dp[i - 1][j - 1];
} else {
if (dp[i - 1][j] == -1 && dp[i - 1][100 - 1] == -1) continue;
if (l >= a[i]) {
dp[i][j] = max(dp[i - 1][j], dp[i - 1][100 - 1]) + 1;
} else {
dp[i][j] = max(dp[i - 1][j], dp[i - 1][100 - 1]);
}
}
}
}
int ans = 0;
for (int i = 100 - y; i <= 100 + x; ++i) {
ans = max(ans, dp[n][i]);
}
//print();
printf("%d\n", ans);
}
return 0;
}
/*
3 6 1 2
1 3 4 5 6 4
0 6 3 1
1 0 1 0 1 0
3 6 1 3
3 4 3 3 4 4
3 6 1 2
3 4 3 4 3 4
3 6 1 2
4 5 4 5 4 5
5 6 1 2
6 5 6 4 5 6
5 6 2 3
6 5 6 5 6 5
4 6 1 6
5 5 5 5 5 5
4 6 5 2
5 5 0 0 5 5
4
6
4
4
2
3
3
1
5
*/
zoj 3911
线段树区间更新，点更新，区间查询。好久不写，不是很会写了 （尴尬 - -#）
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <utility> // pair
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#define pk printf("lalala");
using namespace std;
#define PI acos(-1.0)
#define EXP exp(1.0) // 自然对数
#define ESP 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
typedef long long ll;
const int MAX_N = 10000005;
int prime[MAX_N];
bool is_prime[MAX_N];
int sieve(int n)
{
int p = 0;
for (int i = 0; i <= n; ++i) is_prime[i] = true;
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; ++i) {
if (is_prime[i]) {
prime[p++] = i;
for (int j = 2 * i; j <= n; j += i)
is_prime[j] = false;
}
}
return p;
}
#define lson (o<<1)
#define rson (o<<1|1)
#define mid ((l+r)>>1)
const int N = 100005;
int tr[N * 3];
int ans[N * 3];
int v, yl, yr;
void pushup(int o)
{
ans[o] = ans[lson] + ans[rson];
}
void build(int o, int l, int r)
{
if (l == r) {
scanf("%d", &tr[o]);
if (is_prime[ tr[o] ]) ans[o] = 1;
else ans[o] = 0;
return ;
}
build(lson, l, mid);
build(rson, mid + 1, r);
pushup(o);
}
void pushdown(int o, int l, int r)
{
if (tr[o] == 0) return ;
tr[lson] = tr[rson] = tr[o];
if (is_prime[ tr[o] ]) {
ans[lson] = mid - l + 1;
ans[rson] = r - mid;
} else {
ans[lson] = ans[rson] = 0;
}
tr[o] = 0;
}
void add(int o, int l, int r)
{
if (l == r) {
tr[o] += v;
if (is_prime[ tr[o] ]) ans[o] = 1;
else ans[o] = 0;
return ;
}
pushdown(o, l, r);
if (mid >= yl) add(lson, l, mid);
else add(rson, mid + 1, r);
pushup(o);
}
void update(int o, int l, int r)
{
if (yl <= l && yr >= r) {
tr[o] = v;
if (is_prime[v]) {
ans[o] = r - l + 1;
} else {
ans[o] = 0;
}
return ;
}
pushdown(o, l, r);
if (yl <= mid) update(lson, l, mid);
if (yr > mid) update(rson, mid + 1, r);
pushup(o);
}
int query(int o, int l, int r)
{
if (yl <= l && yr >= r) return ans[o];
pushdown(o, l, r);
int res = 0;
if (yl <= mid) res += query(lson, l, mid);
if (yr > mid) res += query(rson, mid + 1, r);
return res;
}
int main()
{
sieve(10000000);
int t;
scanf("%d", &t);
while (t--) {
int n, q;
scanf("%d%d", &n, &q);
clr(tr, 0);
build(1, 1, n);
while (q--) {
char op[4];
scanf("%s", op);
if (*op == ‘A‘) {
scanf("%d%d", &v, &yl);
add(1, 1, n);
} else if (*op == ‘R‘) {
scanf("%d%d%d", &v, &yl, &yr);
update(1, 1, n);
} else {
scanf("%d%d", &yl, &yr);
printf("%d\n", query(1, 1, n));
}
}
}
return 0;
}
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原文地址：http://www.cnblogs.com/wenruo/p/5161909.html
