c++
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long m,s,t,dp[300001][4],d[300001];
int main()
{
    ios::sync_with_stdio(false); 
    cin>>m>>s>>t;
    dp[0][1]=m;
    for(int i=1;i<=t;i++)
    {
        if(dp[i-1][1]>=10)
        {
            dp[i][1]=dp[i-1][1]-10;
            dp[i][0]=dp[i-1][0]+60;
        }
        else
        {
            dp[i][1]=dp[i-1][1]+4;
            dp[i][0]=dp[i-1][0];
        }
        if(d[i-1]+17>dp[i][0])
        {
            d[i]=d[i-1]+17;
        }
        else
        {
            d[i]=dp[i][0];
        }
        if(d[i]>=s)
        {
            cout<<"Yes"<<endl;
            cout<<i<<endl;
            return 0;
        }
    }
    cout<<"No"<<endl;
    cout<<d[t]<<endl;
    return 0;
} 

ac了

枚举时间，贪心选择。
感谢 冲啊小笼包 的题解。。
~~~
#include <cstdio>
int m,s,t,x,y,i;
int main(){
scanf("%d%d%d",&m,&s,&t);
for(i = 1;i <= t;i++){
x += 17;
if(m >= 10){m -= 10;y += 60;}
else m += 4;
if(x < y) x = y;
if(x >= s){printf("Yes\n%d\n",i);return 0;}
}
(x >= s) ? printf("Yes\n%d\n",i) : printf("No\n%d\n",x);
return 0;
}
~~~

记录信息
评测状态 Accepted
题目 P1431 守望者的逃离
递交时间 2016-06-15 10:59:57
代码语言 C++
评测机 ShadowShore
消耗时间 15 ms
消耗内存 556 KiB
评测时间 2016-06-15 10:59:58
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 552 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 556 KiB, score = 10
Accepted, time = 15 ms, mem = 556 KiB, score = 100
代码
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<string>
using std::swap;
using std::sort;
int f[15],n,d,m,t,k,dist,time;
inline int read()
{
int c=getchar(),temp=0;
for(;c<48||57<c;c=getchar());
do
{
temp=(temp<<3)+(temp<<1)+c-48;
c=getchar();
}while(48<=c&&c<=57);
return temp;
}
inline int max(int x,int y)
{
x-=y;return y+(x&(~(x>>31)));
}
int main()
{
n=read();m=read();t=read();
k=n/10;n%=10;
if(k*60>=m)
{
printf("Yes\n%d",(m+59)/60);
return 0;
}
if(k>=t)
{
printf("No\n%d",t*60);
return 0;
}
m-=k*60;dist+=k*60;
t-=k;time+=k;
k=std::min(m/120,t/7)-1;
k&=~(k>>31);
m-=k*120;dist+=k*120;
t-=k*7;time+=k*7;
for(int i=1;i<=t;++i)
{
if(n>=10) n-=10,d+=60;
else n+=4;
f[i]=max(f[i-1]+17,d);
if(f[i]>=m)
{
printf("Yes\n%d",time+i);
return 0;
}
}
printf("No\n%d",dist+f[t]);
return 0;
}

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
using namespace std;
const int N = 300000;
bool flag = false;
int m,s,t,f[N];
int main(){
scanf("%d%d%d",&m,&s,&t);
for(int i = 1;i <= t;i ++){
if(m >= 10){
f[i] = f[i - 1] + 60;
m -= 10;
}
else {
f[i] = f[i - 1];
m += 4;
}
}
for(int i = 1;i <= t;i ++){
if(f[i] < f[i - 1] + 17)f[i] = f[i - 1] + 17;
if(f[i] >= s){ printf("Yes\n%d",i);break; }
else if(i == t)printf("No\n%d",f[i]);
}
return 0;
}

超简单 清晰的做法！
http://www.ofsxb.com/archives/486

蒟蒻求本题的DP背包做法。。。

记录信息
评测状态 Accepted
题目 P1431 守望者的逃离
递交时间 2015-08-07 21:51:01
代码语言 C++
评测机 VijosEx
消耗时间 90 ms
消耗内存 520 KiB
评测时间 2015-08-07 21:51:02
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 520 KiB, score = 10
测试数据 #1: Accepted, time = 1 ms, mem = 520 KiB, score = 10
测试数据 #2: Accepted, time = 2 ms, mem = 520 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 520 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 520 KiB, score = 10
测试数据 #5: Accepted, time = 36 ms, mem = 520 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 520 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 520 KiB, score = 10
测试数据 #8: Accepted, time = 31 ms, mem = 520 KiB, score = 10
测试数据 #9: Accepted, time = 5 ms, mem = 520 KiB, score = 10
Accepted, time = 90 ms, mem = 520 KiB, score = 100
代码
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int run[35];
int prerun[35];
int main()
{
int m,s,t;
scanf("%d%d%d",&m,&s,&t);
if((s+59)/60<=(m/10))
{
printf("Yes\n");
printf("%d",(s+59)/60);
return 0;
}
int now=m%10;
m/=10;

if(t<m)
{
printf("No\n");
printf("%d",t*60);
return 0;
}
t-=m;
for(int i=0;i<=32;i++)
{
prerun[i]=-100000000;
run[i]=-100000000;
}
prerun[now]=m*60;
for(int i=1;i<=t;i++)
{
for(int j=0;j<=20;j+=1)
{
run[j]=prerun[j]+17;
if(j<=18)
run[j]=max(run[j],prerun[j+10]+60);
if(j>=4)
run[j]=max(run[j],prerun[j-4]);
if(run[j]>=s)
{
printf("Yes\n");
printf("%d",i+m);
return 0;
}
}
memcpy(prerun,run,sizeof(run));
}
int maxans=run[0];
for(int i=1;i<=20;i++)
maxans=max(maxans,run[i]);
printf("No\n%d",maxans);
}

…………………………是Yes不是YES。。。是No不是NO。。。！！！！！我也是跪下了……

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

int a[1000000];

int main()
{
int m,s,t;
cin>>m>>s>>t;
int bri=0;
int i;
for(i=0; i<=t; ){
while(m<10){
m+=4;
i++;
a[i] = a[i-1];
}
i++;
bri+=60;
a[i] = bri;
m-=10;
}
for(i=1; i<=t; i++)
a[i] = max(a[i],a[i-1]+17);
int flg=0;
for(i=1; i<=t; i++)
if(a[i] > s){
flg=1;
break;
}
if(flg == 1)
cout<<"Yes"<<endl<<i;
else
cout<<"No"<<endl<<a[t];
system("pause");
return 0;
}

var
a,b,c,d,e,f:longint;
begin
read(a,b,c);
e:=0;
f:=0;
for d := 1 to c do
begin
e:=e+17;
if a>=10 then
begin
a:=a-10;
f:=f+60;
end
else a:=a+4;
if f>e then e:=f;
if e>=b then
begin
writeln('Yes');
write(d);
halt;
end;
end;
if e>=b then
begin
writeln('Yes');
write(d);
halt;
end;
writeln('No');
write(e);
end.

这个方法应该不算贪心：计算出只用闪烁法术（法力不够就停着不走）时每秒位置

然后再循环，每秒都用走的，若用走的不用如闪烁法术的，就取用闪烁法术的位置

否则取用走的位置

这种方法避免了我的思维障碍（不再需要使用贪心时还考虑休息问题）

#include<stdio.h>
int main( )
{
int g[300001]={0},m,s,t,i,ans,x=0,flag=0;
scanf("%d %d %d",&m,&s,&t);

for(i=1;i<=t;i++)
{
if(m>=10) {m=m-10;g[i]=g[i-1]+60;}
else {m=m+4;g[i]=g[i-1];}

}

for(i=1;i<=t;i++)
{
ans=ans+17;

if(ans<=g[i]) ans=g[i];

if(ans>=s) {printf("Yes\n%d",i);flag=1;break;}
}

if(flag==0) {printf("No\n");printf("%d",ans);}

return 0;
}

贪心1：很显然\({m}\geq{10}\)时一定用闪烁。
贪心2：注意到对于每一个7s,闪烁+休息的距离至少为\(120\),而走路的距离总是\(119\),而一定存在一个\(0\leq a \leq 6\)满足\({a}\equiv {t}\pmod{7}\),所以在t>6 and s>=119时用闪烁+休息。
剩下的部分搜索一下就行了。

Pascal Code

var
    m,s,t,t1,s1:longint;  
//s表示剩下的路程，t表示剩下的时间，s1、t1表示剩下的路程和时间
    bestt,bests:longint;
//bestt表示能逃离时最多剩下的时间，bests表示不能逃离时最少剩下的路程
procedure qmove;
begin
    while m<10 do begin m:=m+4;dec(t);end;
    m:=m-10;
    s:=s-60;
    dec(t);
end;
procedure dfs(m,s,t:longint);  //搜索
begin
  if s<=0 then begin if t>bestt then bestt:=t;exit;end;
  if t=0 then begin if s<bests then bests:=s;exit;end;
  if m>=10 then begin dfs(m-10,s-60,t-1);exit;end;  //贪心1
  dfs(m+4,s,t-1);  //休息
  dfs(m,s-17,t-1);  //走路
end;
begin
  readln(m,s,t);
  t1:=t;
  s1:=s;
  m:=m div 2 *2;
  while (m>=10) and (s>0) and (t>0) do
    qmove;                //贪心1
  if s<=0 then begin writeln('Yes');writeln(t1-t);halt;end;  //能逃离
  if t<=0 then begin writeln('No');writeln(s1-s);halt;end;  //不能逃离
  while (t>6) and (s>=119) do
    qmove;              //贪心2
  bestt:=-maxlongint;
  bests:=maxlongint;
  dfs(m,s,t);
  if bestt>-maxlongint then begin writeln('Yes');writeln(t1-bestt);halt;end;  //能逃离
  if bests<maxlongint then begin writeln('No');writeln(s1-bests);halt;end;  //不能逃离
end.

考虑各种情况模拟即可，时空复杂度都是O(1)

　　贪心，用闪烁的平均速度（跑完后恢复为原来的魔力值）为17.14+，比正常跑地快一点，因此能用闪烁就尽量用，用不了就恢复魔法（岛要没嘞，还不快跑！）。
#include <cmath>
#include <iostream>
using namespace std;

int t, s, m, mt = 1e6, md;

inline int run(int rest) {
if(rest <= 0)
return 0;
return rest / 17 + (rest % 17 ? 1 : 0);
}

int main() {
cin >> m >> s >> t;
for(int i = 0, mp = m, d = 0; i <= t; ++i) {
mt = min(mt, i + run(s - d));
md = max(md, d + 17 * (t - i));
if(mp >= 10)
mp -= 10, d += 60;
else
mp = min(mp + 4, m);
}
if(mt <= t)
cout << "Yes" << endl << mt << endl;
else
cout << "No" << endl << md << endl;
return 0;
}

感谢楼下大神的算法 一开始我想的是模拟 就是各种if各种特判 结果发现特判非常多根本停不下来- -

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 560 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #2: Accepted, time = 15 ms, mem = 560 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 564 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 556 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 556 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 556 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 560 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 556 KiB, score = 10
Accepted, time = 90 ms, mem = 564 KiB, score = 100

#include<iostream>
using namespace std;
int main()
{
long m, s, t, i, a = 0, b = 0;
cin >> m >> s >> t;
for(i = 1; i <= t; i++)
{
a += 17;

if(m >= 10)
{
m -= 10;
b += 60;
}
else m += 4;

if(b > a) a = b;

if(a >= s)
{
cout << "Yes" << endl << i;
return 0;
}
}

if(a >= s)
{
cout << "Yes" << endl << i - 1;
return 0;
}

cout << "No" << endl << a;
return 0;
}

#include<iostream>
using namespace std;
main()
{
long m,s,t,i,a=0,b=0;
cin>>m>>s>>t;
for(i=1;i<=t;i++)
{
a+=17;
if(m>=10)
{
m-=10;
b+=60;
}
else
m+=4;
if(b>a)
a=b;
if(a>=s)
{
cout<<"Yes"<<endl<<i;
return 0;
}
}
if(a>=s)
{
cout<<"Yes"<<endl<<i-1;
return 0;
}
cout<<"No"<<endl<<a;
}

var
a,b,c,d,e,f:longint;
begin
read(a,b,c);
e:=0;
f:=0;
for d := 1 to c do
begin
e:=e+17;
if a>=10 then
begin
a:=a-10;
f:=f+60;
end
else a:=a+4;
if f>e then e:=f;
if e>=b then
begin
writeln('Yes');
write(d);
halt;
end;
end;
if e>=b then
begin
writeln('Yes');
write(d);
halt;
end;
writeln('No');
write(e);
end.

评测结果

编译成功

foo.cpp: In function 'int main()':
foo.cpp:69:14: warning: suggest explicit braces to avoid ambiguous 'else' [-Wparentheses]

测试数据 #0: Accepted, time = 0 ms, mem = 564 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 564 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 564 KiB, score = 10

测试数据 #6: Accepted, time = 15 ms, mem = 568 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 568 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 560 KiB, score = 10

Accepted, time = 15 ms, mem = 568 KiB, score = 100
代码

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int M , S , T;
int t;
int s1 , s2;
int flag;

int main()
{
while( scanf( "%d %d %d" , &M , &S , &T ) != EOF )
{
s1 = s2 = t = 0;
flag = 0;
while( M >= 10 && t < T )
{
M -= 10;
s1 += 60;
s2 += 60;
t++;
if( s1 >= S )
{
flag = 1;
cout << "Yes\n" << t << endl;
break;
}
}
while( t < T && flag != 1 )
{
if( M >= 10 )
{
M -= 10;
s1 += 60;
s2 += 17;
t++;
continue;
}
if( T - t == 2 && M <= 5 )
s1 += 17;
else if( T - t == 1 && M <= 9 )
s1 += 17;
else if( S - s1 <= 34 )
s1 += 17;
else if( M <= 5 && S - s1 <= 51 )
s1 += 17;
else if( M <= 1 && S - s1 <= 68 )
s1 += 17;
else
M += 4;
s2 += 17;
t++;
if( s1 >= S )
{
cout << "Yes\n" << t << endl;
flag = 1;
break;
}
else if( s2 >= S )
{
cout << "Yes\n" << t << endl;
flag = 1;
break;
}
//cout << s1 << " " << s2 << " " << t << endl;
}
if( !flag )
if( s1 >= s2 )
cout << "No\n" << s1 << endl;
else
cout << "No\n" << s2 << endl;
}
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 816 KiB, score = 10
Accepted, time = 0 ms, mem = 816 KiB, score = 100

水炸天~~~貌似普通贪心就秒杀了。CODE：

var m,s,t,i,s1,s2:longint;
begin
read(m,s,t);s1:=0;s2:=0;
for i:=1 to t do begin
s1:=s1+17;
if m>=10 then begin m:=m-10;s2:=s2+60;end else m:=m+4;
if s2>s1 then s1:=s2;
if s1>=s then begin writeln('Yes');writeln(i);halt;end;
end;
if s1>=s then begin writeln('Yes');writeln(i);halt;end;
writeln('No');writeln(s1);
end.