Coder-Strike 2014 - Round 2

t题目链接：Coder-Strike 2014 - Round 2

A题：简单水题，注意能添加重复的数字，因此只要判断能否把Min和Max添加好，就可以了

B题：开一个sum计算每个聊天总和，和一个s计算每个人在每个聊天总和，最后每个人就用总和减掉自己发送的即可

C题：最优策略为先把非特殊的答完，然后从最大的开始答

D题：dp，状态为dp[i][j][k]，i表示当前长度，j表示前面数字的总和，k表示能否组成，然后进行记忆化搜索

代码：

A：

#include <stdio.h>
#include <string.h>

int n, m, Min, Max, num[105];

bool judge() {
int f1 = 0, f2 = 0;
for (int i = 0; i < m; i++) {
if (num[i] < Min || num[i] > Max)
return false;
if (num[i] == Min) f1--;
if (num[i] == Max) f2--;
}
int yu = n - m;
if (!f1) yu--;
if (!f2) yu--;
if (yu < 0) return false;
return true;
}

int main() {
scanf("%d%d%d%d", &n, &m, &Min, &Max);
for (int i = 0; i < m; i++)
scanf("%d", &num[i]);
if (judge()) printf("Correct\n");
else printf("Incorrect\n");
return 0;
}

B：
#include <stdio.h>
#include <string.h>

const int N = 20005;

int n, m, k, vis
[15], sum[15], s
[15], ans
;

int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
scanf("%d", &vis[i][j]);
}
int x, y;
while (k--) {
scanf("%d%d", &x, &y);
x--; y--;
sum[y]++;
s[x][y]++;
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (vis[i][j])
ans[i] += (sum[j] - s[i][j]);
}
}
for (int i = 0; i < n - 1; i++)
printf("%d ", ans[i]);
printf("%d\n", ans[n - 1]);
return 0;
}

C：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 10000005;
int n, m;
long long sum, Max, vis
;
struct Num {
long long num;
int vis;
} num
;

bool cmp(Num a, Num b) {
return a.num > b.num;
}

int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%lld", &num[i].num);
sum += num[i].num;
Max = max(Max, num[i].num);
}
int v;
for (int i = 0; i < m; i++) {
scanf("%d", &v);
sum -= num[v - 1].num;
num[v - 1].vis = 1;
}
sort(num, num + n, cmp);
for (int i = 0; i < n; i++) {
if (num[i].vis) {
if (sum >= num[i].num)
break;
sum += num[i].num;
m--;
}
}
for (int i = 0; i < m; i++)
sum *= 2;
printf("%lld\n", sum);
return 0;
}

D：
#include <stdio.h>
#include <string.h>

const int MOD = 1000000007;
const int N = 2005;
const int M = 8050;
int n, k, num
, dp
[M][2], vis
[M][2];

int solve(int len, int s, int flag) {
if (s>=(1<<k)) flag = 1;
if (vis[len][s][flag])
return dp[len][s][flag];
vis[len][s][flag] = 1;
dp[len][s][flag] = 0;
if (len == n) {
return dp[len][s][flag] = flag;
}
if (num[len] == 0) {
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, s + 2, flag)) % MOD;
if ((s&(1<<1)) == 0) {
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, s + 4, flag)) % MOD;
}
else {
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, 4, flag)) % MOD;
}
}
else {
if (num[len] == 2)
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, s + num[len], flag)) % MOD;
else {
if ((s&(1<<1)) == 0) {
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, s + num[len], flag)) % MOD;
}
else {
dp[len][s][flag] = (dp[len][s][flag] + solve(len + 1, num[len], flag)) % MOD;
}
}
}
return dp[len][s][flag];
}

int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++)
scanf("%d", &num[i]);
printf("%d\n", solve(0, 0, 0));
return 0;
}