A:
#include <cstdio>
#include <cstring>
int main() {
int T;
scanf("%d", &T);
while(T --) {
double x, ans = 0;
for(int i = 0; i < 12;i ++) {
scanf("%lf", &x);
ans += x;
}
ans /= 12;
char b[1000];
sprintf(b,"%.2f", ans);
int len = strlen(b);
if(b[len-1] == '0' && b[len-2] == '0') b[len-3] = '\0';
else if(b[len-1] == '0') b[len-1] = '\0';
printf("$%s\n", b);
}
return 0;
}B:
==明天再上代码，，
C:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define N 505
char mp[N][N];
int n,m;
int dp[N][N][4];
void work(){
//**
//*
for(int i = n-1; i>= 1; i--)
for(int j = m-1; j >= 1; j--)
{
int tmp = 1;
for(int k = 1; i+k<=n && j+k<=m ; k++) {
if(mp[i+k][j]==mp[i][j+k])
tmp++;
else break;
}
dp[i][j][3] = min(tmp, dp[i+1][j+1][3]+2);
}
//**
// *
for(int i = n-1; i >= 1; i--)
for(int j = 2; j <= m; j++)
{
int tmp = 1;
for(int k = 1; i+k<=n && j-k>0 ; k++) {
if(mp[i+k][j]==mp[i][j-k])
tmp++;
else break;
}
dp[i][j][2] = min(tmp, dp[i+1][j-1][2]+2);
}
//*
//**
for(int i = 2; i <= n; i++)
for(int j = m-1; j >= 1; j--)
{
int tmp = 1;
for(int k = 1; i-k>0 && j+k<=m ; k++) {
if(mp[i-k][j]==mp[i][j+k])
tmp++;
else break;
}
dp[i][j][1] = min(tmp, dp[i-1][j+1][1]+2);
}
// *
//**
for(int i = 2; i <= n; i++)
for(int j = 2; j <= m; j++)
{
int tmp = 1;
for(int k = 1; i-k>0 && j-k>0 ; k++) {
if(mp[i-k][j]==mp[i][j-k])
tmp++;
else break;
}
dp[i][j][0] = min(tmp, dp[i-1][j-1][0]+2);
}
}
int main() {
int i,j,k,T;scanf("%d",&T);
while (T--) {
scanf("%d%d",&n,&m);
for(i = 1; i<= n; i++) scanf("%s",mp[i]+1);
memset(dp, 0, sizeof dp);
work();
for(i = 1; i <= n; i++) for(j = 1; j <= m; j++) for(k = 0; k < 4; k++) dp[i][j][k] = max(dp[i][j][k], 1);
int ans = 0;
for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
for(k = 0; k < 4; k++) {
int len = dp[i][j][k];
len = ((len+1)*len)/2;
ans = max(ans, len);
}
cout<<ans<<endl;
}
return 0;
}
D:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <set>
#include <vector>
using namespace std;
typedef int ll;
#define N 100010
ll n, m;
ll f[N];
ll find(ll x){return x==f[x]?x:f[x]=find(f[x]);}
bool Union(ll x, ll y){
ll fx = find(x), fy = find(y);
if(fx==fy)return false;
if(fx>fy)swap(fx,fy);
f[fx] = fy;
return true;
}
struct node{
int from ,to;
}edge[100010];
vector<int>ans;
int main() {
ll u,v,i,j;
while(~scanf("%d %d",&n,&m)) {
ans.clear();
for(i = 0; i < n; i++)f[i] = i;
for(i = 1; i <= m; i++) {
scanf("%d %d",&edge[i].from, &edge[i].to);
}
int now = n;
for(i = m; i >= 1; i--) {
ans.push_back(now);
u = edge[i].from, v = edge[i].to;
if(Union(u,v))
now--;
}
for(i = ans.size()-1; i>=0;i--)
printf("%d\n",ans[i]);
}
return 0;
}E:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <iostream>
#include <algorithm>
#include <vector>
typedef long long ll;
const int N = 100000;
std::vector<int> prime;
void prepare() {
bool vis[N];
for (int i = 2; i < N; ++i)
if (!vis[i]) {
prime.push_back(i);
if ((ll)i * i >= N)
continue;
for (int j = i * i; j < N; j += i)
vis[j] = true;
}
}
int main() {
prepare();
int t, g, l, v, x;
scanf("%d", &t);
while (t -- > 0) {
scanf("%d%d", &g, &l);
if (g == 0 || l % g != 0 || g > l) {
puts("0");
} else {
x = v = l / g;
ll ans = 1;
for (int i = 0; i < prime.size(); ++i) {
if ((ll)prime[i] * prime[i] > x)
break;
if (v % prime[i] == 0) {
int cnt = 0;
while (v % prime[i] == 0) {
v /= prime[i];
++cnt;
}
ans = ans * 6 * cnt;
}
}
if (v != 1) {
ans *= 6;
}
printf("%I64d\n", ans);
}
}
return 0;
}G:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <iostream>
#include <algorithm>
const int N = 6;
int up[N][N], lef[N][N], vis[N][N];
int r, c, ans;
void dfs(int i, int j, int cnt) {
if (i == r && j == c + 1) {
if (cnt > ans)
ans = cnt;
} else if (j == c + 1)
dfs(i + 1, 1, cnt);
else {
//not put
if (!vis[i][j]) {
lef[i][j] = lef[i][j - 1];
up[i][j] = up[i - 1][j];
dfs(i, j + 1, cnt);
}
//put
if (vis[i][j]) {
if (up[i - 1][j] == 1 || up[i - 1][j] == 3)
up[i][j] = 2;
else
up[i][j] = 0;
if (lef[i][j - 1] == 1 || lef[i][j - 1] == 3)
lef[i][j] = 2;
else
lef[i][j] = 0;
dfs(i, j + 1, cnt);
} else {
if (up[i - 1][j] == 2 || lef[i][j - 1] == 2 || up[i - 1][j] == 3 || lef[i][j - 1] == 3)
return ;
if (up[i - 1][j] == 1)
up[i][j] = 3;
else
up[i][j] = 1;
if (lef[i][j - 1] == 1)
lef[i][j] = 3;
else
lef[i][j] = 1;
dfs(i, j + 1, cnt + 1);
}
}
}
void work() {
ans = 0;
memset(lef, 0, sizeof lef);
memset(up, 0, sizeof up);
dfs(1, 1, 0);
printf("%d\n", ans);
}
int main() {
int m, x, y;
while (~scanf("%d%d", &r, &c)) {
memset(vis, 0, sizeof vis);
scanf("%d", &m);
while (m -- > 0) {
scanf("%d%d", &x, &y);
++ x;
++ y;
vis[x][y] = true;
}
work();
}
return 0;
}#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <set>
#include <vector>
using namespace std;
typedef int ll;
#define inf 100000000
#define N 40
ll n, m;
int a[N],b[N],sa[N],sb[N];
int x[N][N][N][N];
int dfs(int al, int ar, int bl, int br){ //先手面对这个局势时能获得的最大价值
if(x[al][ar][bl][br]!=-1)
return x[al][ar][bl][br];
if(al>ar && bl>br)
return 0;
int c[4] = {0};
int A, ans = 0, sum = sa[ar]-sa[al-1] + sb[br]-sb[bl-1];
if(al<=ar) {
c[0] = dfs(al+1, ar, bl, br);
c[1] = dfs(al, ar-1, bl, br);
A = sum - c[0]; ans = max(ans, A);
A = sum - c[1]; ans = max(ans, A);
}
if(bl<=br) {
c[2] = dfs(al, ar, bl+1, br);
c[3] = dfs(al, ar, bl, br-1);
A = sum - c[2]; ans = max(ans, A);
A = sum - c[3]; ans = max(ans, A);
}
x[al][ar][bl][br] = ans;
return ans;
}
int main() {
ll u,v,i,j,T;scanf("%d",&T);
while(T--) {
memset(x, -1, sizeof x);
scanf("%d",&n);
for(i = 1; i <= n; i++)scanf("%d",&a[i]);
for(i = 1; i <= n; i++)scanf("%d",&b[i]);
sa[0] = sb[0] = 0;
for(i = 1; i <= n; i++) {
sa[i] = sa[i-1]+a[i];
sb[i] = sb[i-1]+b[i];
}
// for(i = 1; i <= n; i++)printf("%d ",sa[i]); puts("");
// for(i = 1; i <= n; i++)printf("%d ",sb[i]); puts("");
printf("%d\n",dfs(1,n,1,n));
}
return 0;
}
I:
无码。。
J:
#include<cstdio>
const int mod = 2011;
const int e1 = 1944,e2 = 1609;//30,5的逆元
inline int qPow(int x,int p)
{
int a=6,ans=1;
while(x)
{
if(x&1) ans = ans*a%p;
a = a*a%p;
x >>= 1;
}
return ans;
}
int main()
{
int n,tmp,tmp1,tmp2,ans1,ans2;
while(scanf("%d",&n),n)
{
tmp = qPow(n,mod);
tmp1 = (tmp+24+mod)%mod;
tmp2 = (tmp-1+mod)%mod;
ans1 = tmp1*e1%mod;
ans2 = tmp2*e2%mod;
printf("%d %d\n",ans1,ans2);
}
return 0;
}
2013 通化邀请赛ABCDEGHIJ题解,布布扣,bubuko.com
原文地址：http://blog.csdn.net/qq574857122/article/details/37583911
