Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 78 Accepted Submission(s) : 47
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Problem Description
杞溂鍒頒簡鏀惰幏鐨勫鑺傦紝鐢變簬鏈塗T鐨勪笓涓氭寚瀵鹼紝Lele鑾峰緱浜嗗ぇ涓版敹銆傜壒鍒槸姘存灉錛孡ele涓鍏辯浜哊縐嶆按鏋滐紝鏈夎嫻鏋滐紝姊ㄥ瓙錛岄钑夛紝瑗跨摐……涓嶄絾鍛抽亾濂藉悆錛屾牱瀛愭洿鏄ソ鐪嬨?br>
浜庢槸錛屽緢澶氫漢浠厱鍚嶈屾潵錛屾壘Lele涔版按鏋溿?br>
鐢氳嚦榪炲ぇ鍚嶉紟榧庣殑HDU ACM鎬繪暀澶?lcy 涔熸潵浜嗐俵cy鎶涘嚭涓鎵撶櫨鍏冨ぇ閽烇紝"鎴戣涔扮敱M涓按鏋滅粍鎴愮殑姘存灉鎷肩洏錛屼笉榪囨垜鏈変釜灝忓皬鐨勮姹傦紝瀵逛簬姣忕姘存灉錛屼釜鏁頒笂鎴戞湁闄愬埗錛屾棦涓嶈兘灝戜簬鏌愪釜鐗瑰畾鍊鹼紝涔熶笉鑳藉ぇ浜庢煇涓壒瀹氬箋傝屼笖鎴戜笉瑕佷袱浠戒竴鏍風(fēng)殑鎷肩洏銆備綘闅忔剰鎼厤錛屼綘鑳界粍鍑哄灝戠涓嶅悓鐨勬柟妗堬紝鎴戝氨涔板灝戜喚錛?
鐜板湪灝辮浣犲府甯甃ele錛屽府浠栫畻涓綆楀埌搴曡兘澶熷崠鍑哄灝戜喚姘存灉鎷肩洏緇檒cy浜嗐?br>
娉ㄦ剰錛屾按鏋滄槸浠ヤ釜涓哄熀鏈崟浣嶏紝涓嶈兘澶熷啀鍒嗐傚浜庝袱縐嶆柟妗堬紝濡傛灉鍚勭姘存灉鐨勬暟鐩兘鐩稿悓錛屽垯璁や負(fù)榪欎袱縐嶆柟妗堟槸鐩稿悓鐨勩?br>
鏈緇圠ele鎷夸簡榪欑瑪閽憋紝鍙堝彲浠ョ戶緇粬鐨勫涓氫簡锝?br>
浜庢槸錛屽緢澶氫漢浠厱鍚嶈屾潵錛屾壘Lele涔版按鏋溿?br>
鐢氳嚦榪炲ぇ鍚嶉紟榧庣殑HDU ACM鎬繪暀澶?lcy 涔熸潵浜嗐俵cy鎶涘嚭涓鎵撶櫨鍏冨ぇ閽烇紝"鎴戣涔扮敱M涓按鏋滅粍鎴愮殑姘存灉鎷肩洏錛屼笉榪囨垜鏈変釜灝忓皬鐨勮姹傦紝瀵逛簬姣忕姘存灉錛屼釜鏁頒笂鎴戞湁闄愬埗錛屾棦涓嶈兘灝戜簬鏌愪釜鐗瑰畾鍊鹼紝涔熶笉鑳藉ぇ浜庢煇涓壒瀹氬箋傝屼笖鎴戜笉瑕佷袱浠戒竴鏍風(fēng)殑鎷肩洏銆備綘闅忔剰鎼厤錛屼綘鑳界粍鍑哄灝戠涓嶅悓鐨勬柟妗堬紝鎴戝氨涔板灝戜喚錛?
鐜板湪灝辮浣犲府甯甃ele錛屽府浠栫畻涓綆楀埌搴曡兘澶熷崠鍑哄灝戜喚姘存灉鎷肩洏緇檒cy浜嗐?br>
娉ㄦ剰錛屾按鏋滄槸浠ヤ釜涓哄熀鏈崟浣嶏紝涓嶈兘澶熷啀鍒嗐傚浜庝袱縐嶆柟妗堬紝濡傛灉鍚勭姘存灉鐨勬暟鐩兘鐩稿悓錛屽垯璁や負(fù)榪欎袱縐嶆柟妗堟槸鐩稿悓鐨勩?br>
鏈緇圠ele鎷夸簡榪欑瑪閽憋紝鍙堝彲浠ョ戶緇粬鐨勫涓氫簡锝?br>
Input
鏈鐩寘鍚緇勬祴璇曪紝璇峰鐞嗗埌鏂囦歡緇撴潫(EOF)銆?br>姣忕粍嫻嬭瘯絎竴琛屽寘鎷袱涓鏁存暟N鍜孧(鍚箟瑙侀鐩弿榪幫紝0<N,M<=100)
鎺ヤ笅鏉ユ湁N琛屾按鏋滅殑淇℃伅錛屾瘡琛屼袱涓暣鏁癆,B(0<=A<=B<=100)錛岃〃紺鴻嚦灝戣涔拌姘存灉A涓紝鑷沖鍙兘涔拌姘存灉B涓?br>
鎺ヤ笅鏉ユ湁N琛屾按鏋滅殑淇℃伅錛屾瘡琛屼袱涓暣鏁癆,B(0<=A<=B<=100)錛岃〃紺鴻嚦灝戣涔拌姘存灉A涓紝鑷沖鍙兘涔拌姘存灉B涓?br>
Output
瀵逛簬姣忕粍嫻嬭瘯錛屽湪涓琛岄噷杈撳嚭鎬誨叡鑳藉鍗栫殑鏂規(guī)鏁般?br>棰樼洰鏁版嵁淇濊瘉榪欎釜絳旀灝忎簬10^9
Sample Input
2 3 1 2 1 2 3 5 0 3 0 3 0 3
Sample Output
2 12
Author
Linle
Source
ACM紼嬪簭璁捐鏈熸湯鑰冭瘯鈥斺?008-01-02錛? 鏁?17錛?
1
#include<stdio.h>
2int main()
3
{
4
int i,n,m,j,k,a[101],b[101],c1[101],c2[101];
5 while(scanf("%d %d",&n,&m)!=EOF)
6
{
7 for(i=0;i<n;i++)
8 scanf("%d %d",&a[i],&b[i]);
9 for(i=0;i<=m;i++)
10 {
11 c1[i]=0;
12 c2[i]=0;
13
}
14 c1[0]=1;//絎竴嬈＄粨鏉熶互鍚?nbsp;k鐨勫氨琚祴鍊兼垜涓?浜?nbsp;鍙叾浣欒繕涓? 灝辯畻c1[j]涓鐢ㄥ埌榪樻槸娌℃晥鏋?nbsp;鍙湁1鐨刢1[j]鎵嶄細(xì)鏈夋晥鏋?/span>
15 for(i=0;i<n;i++)
16 {
17 for(j=0;j<=m;j++)
18 for(k=a[i];k+j<=m&&k<=b[i];k++)
19 c2[j+k]+=c1[j];
20 for(j=0;j<=m;j++)
21 {
22 c1[j]=c2[j];
23 c2[j]=0;
24 }
25 }
26 printf("%d\n",c1[m]);
27 }
28 return 0;
29
}
#include<stdio.h>2
int main()3
{4
int i,n,m,j,k,a[101],b[101],c1[101],c2[101];5
while(scanf("%d %d",&n,&m)!=EOF)6
{7
for(i=0;i<n;i++)8
scanf("%d %d",&a[i],&b[i]);9
for(i=0;i<=m;i++)10
{11
c1[i]=0;12
c2[i]=0;13
}14
c1[0]=1;//絎竴嬈＄粨鏉熶互鍚?nbsp;k鐨勫氨琚祴鍊兼垜涓?浜?nbsp;鍙叾浣欒繕涓? 灝辯畻c1[j]涓鐢ㄥ埌榪樻槸娌℃晥鏋?nbsp;鍙湁1鐨刢1[j]鎵嶄細(xì)鏈夋晥鏋?/span>15
for(i=0;i<n;i++)16
{17
for(j=0;j<=m;j++)18
for(k=a[i];k+j<=m&&k<=b[i];k++)19
c2[j+k]+=c1[j];20
for(j=0;j<=m;j++)21
{22
c1[j]=c2[j];23
c2[j]=0;24
}25
}26
printf("%d\n",c1[m]);27
}28
return 0;29
}]]>
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 132 Accepted Submission(s) : 69
Font: Times New Roman | Verdana | Georgia
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Problem Description
We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
“Oh, God! How terrible! ”
Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (紜竵) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!
“Oh, God! How terrible! ”
Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (紜竵) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!
Input
Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.
Output
Output the minimum positive value that one cannot pay with given coins, one line for one case.
Sample Input
1 1 3 0 0 0
Sample Output
4
Author
lcy
#include<stdio.h>#include<string.h>int main(){ int i,j,k,a,b,c,d[10001]; while(scanf("%d %d %d",&a,&b,&c)!=EOF) { if(0==a&&0==b&&0==c) break; memset(d,0,sizeof(d)); for(j=0;j<=a;j++) for(k=0;k+j<=a+2*b;k=k+2) d[k+j]++; for(j=0;j<=a+2*b&&d[j];j++)//&&a[j]鏄帓闄や粠0鍒癮+2b涔嬪唴涓嶅瓨鍦ㄧ殑欏?/span> for(k=0;k+j<=a+2*b+5*c;k=k+5) d[k+j]++; for(i=0;i<=a+2*b+5*c+1;i++)//+1鏄洜涓哄鏋滄儏鍐典腑鐨勫叏婊¤凍 閭ｄ箞澶т簬鑼冨洿鐨勭涓涓氨鏄墍姹?/span> { if(d[i]==0) { printf("%d\n",i); break; } } } return 0;}]]>
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