Kill the monster
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Kill the monster
Problem Description
There is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it.
Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.
Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.
Input
The input contains multiple test cases.
Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).
Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).
Output
For each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1.
#include<stdio.h>int m,n,minn;int a[10],b[10],num[10];int dfs(int jishu,int hp){ if(jishu>m) return 0; if(hp<=0) { if(minn>jishu) minn=jishu; return 0; } for(int i=0;i<m;i++) { if(num[i]==0) { num[i]=1; if(hp<=b[i]) { dfs(jishu+1,hp-2*a[i]); } else dfs(jishu+1,hp-a[i]); num[i]=0; } } return 0;}int main(){ while(~scanf("%d%d",&m,&n)) { for(int i=0;i<m;i++) { scanf("%d%d",&a[i],&b[i]); num[i]=0; } minn=12; dfs(0,n); if(minn==12) { printf("-1\n"); } else { printf("%d\n",minn); } }}
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