简单模拟。规律的话就是密匙中A代表加0，B代表加1，以此类推。于是先把密匙大小写同一，然后与A减，得到的再用密文减即可。
需要注意的一点就是密文大小写一开始就定下来了，如果减过了‘a’或者‘A’的话，依照表的规律的话要改成‘Z’或‘z’减掉溢出。
#include<bits/stdc++.h>
using namespace std;
#define RG register
#define ll long long
#define in(x) x = read()
#define lin(x) scanf("%lld",&x)
#define llin(x) x = lread
#define din(x) scanf("%lf",&x)
#define dout(x) printf("%lf",x)
#define out(x) print(x)
#define lout(x) printf("%lld",x)
#define ex putchar('\n')
#define ko putchar(' ')
const int MAXN = 1e4 + 5;
int klen,clen;
string k,c;
inline int read()
{
int X=0,w=1;
char ch=getchar();
while(ch<'0' || ch>'9')
{
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0' && ch<='9') X=(X<<3)+(X<<1)+ch-'0',ch=getchar();
return X*w;
}
inline ll lread()
{
ll X=0,w=1;
char ch=getchar();
while(ch<'0' || ch>'9')
{
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0' && ch<='9') X=(X<<3)+(X<<1)+ch-'0',ch=getchar();
return X*w;
}
void print(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
print(x/10);
}
putchar(x%10+'0');
}
int main()
{
// freopen("vigenere.in","r",stdin);
// freopen("vigenere.out","w",stdout);
cin >> k >> c;
klen = k.size();
clen = c.size();
for(int i = 0;i < klen;i++)
if(isupper(k[i])) k[i] = tolower(k[i]);
int now = 0;
for(int i = 0;i < clen;i++)
{
bool up = 0;
if(isupper(c[i])) up = 1;
if(now == klen) now = 0;
c[i] = char((int)c[i] - ((int)k[now++] - (int)'a'));
if(up)
{
if(c[i] < 'A') c[i] = char((int)'Z' - ((int)'A' - (int)c[i])+1);
}
else
if(c[i] < 'a') c[i] = char((int)'z' - ((int)'a' - (int)c[i])+1);
}
cout << c;
return 0;
}
首先我们分析一下，如果i和j两个相邻那么i排在j前面的必要条件是
total*a[i]/b[j]<total*a[j]/b[i]
也就是说 a[i]*b[i]<a[j]*b[j]
于是我们相乘排个序就好，注意数据较大要用高精度。
#include<bits/stdc++.h>
using namespace std;
#define RG register
#define ll long long
#define in(x) x = read()
#define lin(x) scanf("%lld",&x)
#define llin(x) x = lread
#define din(x) scanf("%lf",&x)
#define dout(x) printf("%lf",x)
#define out(x) print(x)
#define lout(x) printf("%lld",x)
#define ex putchar('\n')
#define ko putchar(' ')
const int MAXN = 1e3 + 5;
int n;
ll kl,kr;
ll l[MAXN],r[MAXN];
int temp[1000000];
string mul,ans;
struct game
{
int l,r;
ll m;
bool operator < (const game x) const
{
return m < x.m;
}
}g[MAXN];
string Mul(string a,ll b)
{
string t;
int la = a.size();
memset(temp,0,sizeof temp);
for(int i = la-1;i >= 0;i--) temp[la-1-i] = a[i] - '0';
ll w = 0;
for(int i = 0;i < la;i++)
{
temp[i] = temp[i]*b + w;
w = temp[i]/10;
temp[i] = temp[i] % 10;
}
while(w) temp[la++] = w % 10,w /= 10;
la--;
while(la >= 0) t += temp[la--] + '0';
return t;
}
string Div(string a,int b)
{
string t;
int d = 0;
if(a == "0") return a;
for(int i=0;i<a.size();i++) {
t += (d * 10 + a[i] - '0') / b + '0';
d = (d * 10 + (a[i] - '0')) % b;
}
int p = 0;
for(int i=0;i<t.size();i++)
if(t[i] != '0') {
p = i;
break;
}
return t.substr(p);
}
inline int read()
{
int X=0,w=1;
char ch=getchar();
while(ch<'0' || ch>'9')
{
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0' && ch<='9') X=(X<<3)+(X<<1)+ch-'0',ch=getchar();
return X*w;
}
inline ll lread()
{
ll X=0,w=1;
char ch=getchar();
while(ch<'0' || ch>'9')
{
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0' && ch<='9') X=(X<<3)+(X<<1)+ch-'0',ch=getchar();
return X*w;
}
void print(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
print(x/10);
}
putchar(x%10+'0');
}
int main()
{
// freopen("game.in","r",stdin);
// freopen("game.out","w",stdout);
in(n); cin >> mul; lin(kr);
for(int i = 1;i <= n;i++)
{
in(g[i].l);in(g[i].r);
g[i].m = g[i].l * g[i].r;
}
sort(g+1,g+1+n);
ans = Div(mul,g[1].r);
mul = Mul(mul,g[1].l);
for(int i = 2;i <= n;i++)
{
// ans = max(sum/g[i].r,ans);
// sum *= g[i].l;
string temp;
temp = Div(mul,g[i].r);
mul = Mul(mul,g[i].l);
// if(ans.size() > temp.size()) ;
if(temp.size() > ans.size())
ans = temp;
else if(temp > ans)
ans = temp;
}
// lout(ans);
cout << ans;
return 0;
}
第三题并不会做需填坑。