给定两个染色区间的端点L_1,R_1,L_2,R_2,求同时染上两种颜色的区间长度
res[i]
记录被染色的情况#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int res[N];
int main(){
int a, b, c, d;
cin >> a >> b >> c >> d;
for(int i = a; i <= b; i ++) res[i]++;
for(int i = c; i <= d; i ++) res[i]++;
int cnt = 0;
for(int i = 0; i <= 200; i ++) if(res[i] == 2) cnt++;
if(cnt) cout << cnt - 1 << endl;
else cout << 0 << endl;
return 0;
}
给定N\times N的表,单元A_{i,j}记录了+i与j的对局输赢平的三种情况,若记录有冲突说明该表不正确
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
char mp[N][N];
int n;
int main(){
cin >> n;
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j++) cin >> mp[i][j];
}
bool flag = 1;
for(int i = 1; i <= n; i ++){
for(int j = i + 1; j <= n; j ++){
if(mp[i][j] == 'D' && mp[j][i] != 'D'){
flag = 0;
break;
}
else if(mp[i][j] == 'W' && mp[j][i] != 'L'){
flag = 0;
break;
}
else if(mp[i][j] == 'L' && mp[j][i] != 'W'){
flag = 0;
break;
}
}
if(!flag) break;
}
if(flag) cout << "correct" << endl;
else cout << "incorrect" << endl;
return 0;
}
给定N个string类型的S串,按照给出S的顺序,对于总共出现过m次S_i,若当前的S_i第一次出现,输出S_i,否则输出S_i(u-1),u表示第几次出现
s[N]
,使得a[s[i]]++
,若a[s[i]] - 1 == 0
说明是第一次出现#include <bits/stdc++.h>
using namespace std;
map<string,int> a;
int n;
const int N = 1e6 + 3;
string s[N];
int main(){
cin >> n;
for(int i = 0; i < n; i ++) cin >> s[i];
for(int i = 0; i < n; i ++){
a[s[i]]++;
int t = a[s[i]] - 1;
if(t != 0){
cout << s[i] << '(' << t << ')' << endl;
}
else cout << s[i] << endl;
}
return 0;
}
投掷硬币为正面,计数器增加,反之计数器清零,给定N次投掷硬币为正面得到的钱X_i,给定M个奖励规则,若计数器的数值达到C_i,将获得Y_i的奖励,求如何使得得到的钱最多
dp[i][j]
表示前i
次投掷,当前计数器值为j
时得到的钱dp[i][j] = dp[i - 1][j - 1] + a[i] + b[j]
dp[i][0] = max(dp[i][0],dp[i - 1][j])
long long
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL n,m;
const LL N = 5010;
LL dp[N][N];
LL a[N];
LL b[N];
int main(){
cin >> n >> m;
for(int i = 1; i <= n; i ++) cin >> a[i];
for(int i = 0; i < m; i ++){
int x, y;
cin >> x >> y;
b[x] = y;
}
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= i; j ++){
dp[i][j] = dp[i - 1][j - 1] + a[i] + b[j];
}
for(int j = 0; j < i; j ++){
dp[i][0] = max(dp[i][0],dp[i - 1][j]);
}
}
LL ans = 0;
for(int i = 0; i <= n; i ++) ans = max(ans,dp[n][i]);
cout << ans << endl;
return 0;
}